maths-trigonometry Mixed 2

Mixed|MATHS

यदि \(\cot^2 a + \tan^2 a = 2\) जहां \(0^\circ \le a \le 90^\circ\), तो \(a\) का मान ज्ञात कीजिए।

• \(0^\circ\)
• \(45^\circ\)
• \(60^\circ\)
• \(90^\circ\)

Explanation:

• \(\cot^2 a+\tan^2 a = 2\)
  \(\Rightarrow \cot^2 a+\tan^2 a-2\cot a\tan a = 0\) (क्योंकि \(\cot a\tan a = 1\))
  \(\Rightarrow (\cot a-\tan a)^2 = 0 \Rightarrow \cot a = \tan a\)
  \(\Rightarrow \tan^2 a = 1 \Rightarrow \tan a = 1 \Rightarrow a = 45^\circ\)

यदि \(\tan^2 A – 6\tan A + 9 = 0,\ 0 < A < 90^\circ\), तो \(6\cot A + 8\sqrt{10}\cos A\) का मान ज्ञात करें।

• \(10\sqrt{10}\)
• \(14\)
• \(10\)
• \(8\sqrt{10}\)

Explanation:

• \(\tan^2 A-6\tan A+9 = 0 \Rightarrow (\tan A-3)^2 = 0 \Rightarrow \tan A = 3\)
  \(\cot A = \frac{1}{3}\), \(\cos A = \frac{1}{\sqrt{10}}\) (आधार 1, लंब 3, कर्ण \(\sqrt{10}\))
  \(6\cot A+8\sqrt{10}\cos A = 6\times \frac{1}{3}+8\sqrt{10}\times \frac{1}{\sqrt{10}} = 2+8 = 10\)

यदि \(0^\circ < \theta < 90^\circ\) है, तो \(\left(\dfrac{1-\cot\theta}{1-\tan\theta}\right)^2 + 1\) का मान ज्ञात कीजिए।

• \(\csc^2\theta\)
• \(\sin^2\theta\)
• \(\cos^2\theta\)
• \(\sec^2\theta\)

Explanation:

• \(\frac{1-\cot\theta}{1-\tan\theta} = \frac{1-\frac{1}{\tan\theta}}{1-\tan\theta} = \frac{\frac{\tan\theta-1}{\tan\theta}}{1-\tan\theta} = -\frac{1}{\tan\theta} = -\cot\theta\)
  \(\left(-\cot\theta\right)^2+1 = \cot^2\theta+1 = \csc^2\theta\)

\(\dfrac{5\cos^2 62^\circ + 5\cos^2 28^\circ – 21}{7\sin^2 35^\circ + 7\sin^2 55^\circ + 1} = ?\)

• \(2\)
• \(-2\)
• \(3\)
• \(-3\)

Explanation:

• \(\cos 62^\circ = \sin 28^\circ\) \(\Rightarrow \cos^2 62^\circ+\cos^2 28^\circ = \sin^2 28^\circ+\cos^2 28^\circ = 1\)
  \(\sin 55^\circ = \cos 35^\circ \Rightarrow \sin^2 35^\circ+\sin^2 55^\circ = 1\)
  व्यंजक \(= \frac{5(1)-21}{7(1)+1} = \frac{-16}{8} = -2\)

यदि \(\dfrac{\sin^2\theta – 3\sin\theta + 2}{\cos^2\theta} = 1\) है, जहाँ \(0^\circ < \theta < 90^\circ\), तो \((\cos 2\theta – \sin 3\theta + \csc 2\theta)\) का मान क्या है?

• \(\dfrac{-3-4\sqrt{3}}{6}\)
• \(\dfrac{-3+4\sqrt{3}}{6}\)
• \(\dfrac{3+4\sqrt{3}}{6}\)
• \(\dfrac{3-4\sqrt{3}}{6}\)

Explanation:

• \(\frac{\sin^2\theta-3\sin\theta+2}{\cos^2\theta} = 1 \Rightarrow \sin^2\theta-3\sin\theta+2 = 1-\sin^2\theta\)
  \(2\sin^2\theta-3\sin\theta+1 = 0 \Rightarrow (2\sin\theta-1)(\sin\theta-1) = 0\)
  \(\sin\theta = \frac{1}{2}\) (\(\sin\theta = 1\) पर \(\theta = 90^\circ\) मान्य नहीं) \(\Rightarrow \theta = 30^\circ\)
  \(\cos 60^\circ-\sin 90^\circ+\csc 60^\circ = \frac{1}{2}-1+\frac{2}{\sqrt{3}} = -\frac{1}{2}+\frac{2\sqrt{3}}{3} = \frac{-3+4\sqrt{3}}{6}\)

यदि \(\sec^2\alpha + 4\cos^2\alpha = 4\) और \(0 \le \alpha \le 90^\circ\) है, तो \(\alpha\) का मान ज्ञात कीजिए।

• \(30^\circ\)
• \(45^\circ\)
• \(60^\circ\)
• \(75^\circ\)

Explanation:

• \(\cos^2\alpha = x\) लें: \(\sec^2\alpha+4\cos^2\alpha = 4 \Rightarrow \frac{1}{x}+4x = 4\)
  \(4x^2-4x+1 = 0 \Rightarrow (2x-1)^2 = 0 \Rightarrow x = \frac{1}{2}\)
  \(\cos^2\alpha = \frac{1}{2} \Rightarrow \cos\alpha = \frac{1}{\sqrt{2}} \Rightarrow \alpha = 45^\circ\)

\(\tan^2\theta + \cot^2\theta + \sec^2\theta – \csc^2\theta\)

• \(2\)
• \(+2\)
• \(0\)
• \(1/2\)

Explanation:

• \(\sec^2\theta = 1+\tan^2\theta\) और \(\csc^2\theta = 1+\cot^2\theta\)
  \(\sec^2\theta+\csc^2\theta-\tan^2\theta-\cot^2\theta = (1+\tan^2\theta)+(1+\cot^2\theta)-\tan^2\theta-\cot^2\theta = 2\)
  नोट: जैसा छपा है (\(\tan^2\theta+\cot^2\theta+\sec^2\theta-\csc^2\theta\)) वह अचर नहीं है (यह \(2\tan^2\theta\) बनता है); उत्तर कुंजी (2) व्यंजक \(\sec^2\theta+\csc^2\theta-\tan^2\theta-\cot^2\theta\) से मेल खाती है।

यदि \(\tan\theta + \sec\theta = 7\) है और \(\theta\) न्यून कोण हो, तो \(5\sin\theta\) का मान होगा-

• \(\dfrac{25}{24}\)
• \(\dfrac{24}{25}\)
• \(\dfrac{1}{24}\)
• \(\dfrac{24}{5}\)

Explanation:

• \(\sec\theta+\tan\theta = 7\) …(i)
  \(\sec^2\theta-\tan^2\theta = 1 \Rightarrow \sec\theta-\tan\theta = \frac{1}{7}\) …(ii)
  (i)\(-\)(ii): \(2\tan\theta = 7-\frac{1}{7} = \frac{48}{7} \Rightarrow \tan\theta = \frac{24}{7}\)
  (i)\(+\)(ii): \(2\sec\theta = \frac{50}{7} \Rightarrow \sec\theta = \frac{25}{7}\)
  \(\sin\theta = \frac{\tan\theta}{\sec\theta} = \frac{24}{25}\)
  \(5\sin\theta = 5\times \frac{24}{25} = \frac{24}{5}\)

यदि \(\tan^2\theta = 1 – a^2\) है, तो \(\sec\theta + \tan^3\theta\csc\theta\) का मान क्या है?

• \((2-a)^{3/2}\)
• \((a^2-1)^{3/2}\)
• \((2-a^2)^{3/2}\)
• \(a^{3/2}\)

Explanation:

• \(\sec\theta+\tan^3\theta\csc\theta = \sec\theta+\tan^2\theta\cdot\frac{\tan\theta}{\sin\theta} = \sec\theta+\tan^2\theta\sec\theta\)
  \(= \sec\theta(1+\tan^2\theta) = \sec^3\theta = (1+\tan^2\theta)^{3/2}\)
  \(= (1+1-a^2)^{3/2} = (2-a^2)^{3/2}\)

यदि \(\cot x – \tan x = \dfrac{3}{2}\) है, तो \(\cot x + \tan x\) का मान क्या होगा?

• \(3\)
• \(2\)
• \(\dfrac{5}{2}\)
• \(\dfrac{7}{2}\)

Explanation:

• \((\cot x+\tan x)^2 = (\cot x-\tan x)^2+4\cot x\tan x = \frac{9}{4}+4 = \frac{25}{4}\)
  \(\cot x+\tan x = \frac{5}{2}\)

यदि \(\sec^2 A + \tan^2 A = \dfrac{4}{17}\) है, तो \(\sec^4 A – \tan^4 A\) का मान क्या होगा?

• \(\dfrac{13}{17}\)
• \(\dfrac{4}{13}\)
• \(\dfrac{4}{17}\)
• \(\dfrac{5}{17}\)

Explanation:

• \(\sec^4 A-\tan^4 A = (\sec^2 A-\tan^2 A)(\sec^2 A+\tan^2 A)\)
  \(= 1\times \frac{4}{17} = \frac{4}{17}\)

यदि \(\sin\theta + \cos\theta = \sqrt{5}\sin(90^\circ – \theta)\) है, तो \(\cot\theta\) का मान ज्ञात कीजिए।

• \(\dfrac{\sqrt{5}-1}{5}\)
• \(\dfrac{\sqrt{5}+1}{4}\)
• \(\dfrac{\sqrt{5}-1}{4}\)
• \(\dfrac{\sqrt{5}+1}{3}\)

Explanation:

• \(\sin(90^\circ-\theta) = \cos\theta\) \(\Rightarrow \sin\theta+\cos\theta = \sqrt{5}\cos\theta\)
  \(\cos\theta\) से भाग देने पर: \(\tan\theta+1 = \sqrt{5} \Rightarrow \tan\theta = \sqrt{5}-1\)
  \(\cot\theta = \frac{1}{\sqrt{5}-1} = \frac{\sqrt{5}+1}{(\sqrt{5}-1)(\sqrt{5}+1)} = \frac{\sqrt{5}+1}{4}\)

यदि \(\csc A – \cot A = \dfrac{1}{4}\) है, तो \(\tan A\) का मान क्या है?

• \(\dfrac{8}{15}\)
• \(\dfrac{8}{17}\)
• \(\dfrac{15}{17}\)
• \(\dfrac{17}{15}\)

Explanation:

• \(\csc A-\cot A = \frac{1}{4}\) \(\Rightarrow \csc A+\cot A = 4\) (क्योंकि \(\csc^2 A-\cot^2 A = 1\))
  जोड़ने पर \(2\csc A = \frac{17}{4} \Rightarrow \csc A = \frac{17}{8}\)
  घटाने पर \(2\cot A = 4-\frac{1}{4} = \frac{15}{4} \Rightarrow \cot A = \frac{15}{8}\)
  \(\tan A = \frac{8}{15}\)

\(\dfrac{\cos A}{1+\tan A} – \dfrac{\sin A}{1+\cot A} = ?\)

• \(\tan A\)
• \(\cos A – \sin A\)
• \(\cos A \cdot \sin A\)
• \(\cos A + \sin A\)

Explanation:

• \(\frac{\cos A}{1+\tan A} = \frac{\cos A}{1+\frac{\sin A}{\cos A}} = \frac{\cos^2 A}{\cos A+\sin A}\)
  \(\frac{\sin A}{1+\cot A} = \frac{\sin^2 A}{\sin A+\cos A}\)
  अंतर \(= \frac{\cos^2 A-\sin^2 A}{\cos A+\sin A} = \frac{(\cos A-\sin A)(\cos A+\sin A)}{\cos A+\sin A} = \cos A-\sin A\)

यदि \(\theta\) एक न्यून कोण और \(\tan\theta + \cot\theta = 2\) है, तो \(\tan^2\theta + \cot^2\theta + 2\tan^5\theta\cot^4\theta\) का मान क्या है?

• \(3\)
• \(4\)
• \(1\)
• \(2\)

Explanation:

• \(\tan\theta+\cot\theta = 2\) \(\Rightarrow \tan\theta+\frac{1}{\tan\theta} = 2 \Rightarrow (\tan\theta-1)^2 = 0 \Rightarrow \tan\theta = 1\), \(\cot\theta = 1\)
  मान \(= 1+1+2\times 1\times 1 = 4\)

यदि \(\sec\theta – 2\cos\theta = \dfrac{7}{2}\), जहाँ \(\theta\) एक धनात्मक न्यून कोण है, तो \(\sec\theta\) का मान ज्ञात कीजिए।

• \(6\)
• \(8\)
• \(5\)
• \(4\)

Explanation:

• माना \(\sec\theta = s\); \(\cos\theta = \frac{1}{s}\)
  \(s-\frac{2}{s} = \frac{7}{2} \Rightarrow 2s^2-7s-4 = 0 \Rightarrow (2s+1)(s-4) = 0\)
  न्यून कोण के लिए \(s > 0\), इसलिए \(\sec\theta = 4\)

यदि \(\tan^2 a = 3 + Q^2\) है, तो \(\sec a + \tan^3 a\csc a = ?\)

• \((3+Q^2)^{3/2}\)
• \((7+Q^2)^{3/2}\)
• \((5-Q^2)^{3/2}\)
• \((4+Q^2)^{3/2}\)

Explanation:

• \(\sec a+\tan^3 a\csc a = \sec a+\tan^2 a\cdot\frac{\tan a}{\sin a} = \sec a+\tan^2 a\sec a = \sec a(1+\tan^2 a) = \sec^3 a\)
  \(\sec^2 a = 1+\tan^2 a = 1+3+Q^2 = 4+Q^2\)
  \(\sec^3 a = (4+Q^2)^{3/2}\)

यदि \(1 + \sin\theta = m\cos\theta\) है, तो \(\sin\theta\) का मान क्या होगा?

• \(\dfrac{2m^2-1}{m^2+1}\)
• \(\dfrac{m^2-1}{m^2+1}\)
• \(\dfrac{m^2+1}{2m^2-1}\)
• \(\dfrac{2m^2-1}{m^2-1}\)

Explanation:

• \(m = \frac{1+\sin\theta}{\cos\theta} \Rightarrow m^2 = \frac{(1+\sin\theta)^2}{1-\sin^2\theta} = \frac{(1+\sin\theta)^2}{(1-\sin\theta)(1+\sin\theta)} = \frac{1+\sin\theta}{1-\sin\theta}\)
  \(m^2(1-\sin\theta) = 1+\sin\theta \Rightarrow m^2-1 = \sin\theta(m^2+1)\)
  \(\sin\theta = \frac{m^2-1}{m^2+1}\)

\(\cos(30^\circ + \theta) – \sin(60^\circ – \theta) = \_\_\_\_\_\)

• \(\dfrac{\sqrt{3}}{2}\)
• \(0\)
• \(\dfrac{1}{2}\)
• \(\dfrac{1}{\sqrt{2}}\)

Explanation:

• \(\sin(60^\circ-\theta) = \cos\{90^\circ-(60^\circ-\theta)\} = \cos(30^\circ+\theta)\)
  इसलिए \(\cos(30^\circ+\theta)-\sin(60^\circ-\theta) = 0\)

यदि \(\tan\theta + \cot\theta = 12\) है, तो \(\tan^2\theta + \cot^2\theta\) का मान क्या होगा?

• \(66\)
• \(128\)
• \(142\)
• \(132\)

Explanation:

• \(\tan^2\theta+\cot^2\theta = (\tan\theta+\cot\theta)^2-2\tan\theta\cot\theta\)
  \(= 12^2-2 = 144-2 = 142\)

यदि \(\sec x – \cos x = 4\) है, तो \(\dfrac{(1+\cos^2 x)}{\cos x}\) का मान क्या होगा?

• \(\dfrac{9}{4}\)
• \(\dfrac{1}{4}\)
• \(2\sqrt{5}\)
• \(\sqrt{5}\)

Explanation:

• \(\sec x-\cos x = 4\) \(\Rightarrow\left(\sec x+\cos x\right)^2 = (\sec x-\cos x)^2+4\sec x\cos x = 16+4 = 20\)
  \(\frac{1+\cos^2 x}{\cos x} = \sec x+\cos x = \sqrt{20} = 2\sqrt{5}\)

\(\dfrac{\sin^2 63^\circ + \sin^2 27^\circ}{\cos^2 17^\circ + \cos^2 73^\circ} = ?\)

• \(0\)
• \(3\)
• \(1\)
• \(2\)

Explanation:

• \(\sin 63^\circ = \cos 27^\circ \Rightarrow \sin^2 63^\circ+\sin^2 27^\circ = \cos^2 27^\circ+\sin^2 27^\circ = 1\)
  \(\cos 73^\circ = \sin 17^\circ \Rightarrow \cos^2 17^\circ+\cos^2 73^\circ = \cos^2 17^\circ+\sin^2 17^\circ = 1\)
  मान \(= \frac{1}{1} = 1\)

यदि \(\tan A\tan B + \dfrac{\cos x}{\cos A\cos B} = 1\) है, तो \(x = ?\)

• \(B\)
• \(A\)
• \(A + B\)
• \(A – B\)

Explanation:

• \(\tan A\tan B+\frac{\cos x}{\cos A\cos B} = 1\)
  \(\cos A\cos B\) से गुणा करने पर: \(\sin A\sin B+\cos x = \cos A\cos B\)
  \(\cos x = \cos A\cos B-\sin A\sin B = \cos(A+B)\)
  \(x = A+B\)

यदि \(\sin^2\theta\cos^2\theta = \dfrac{2}{9}\) है, तो \(\csc^2\theta + \sec^2\theta\) का मान क्या होगा?

• \(\dfrac{7}{2}\)
• \(\dfrac{5}{2}\)
• \(\dfrac{9}{2}\)
• \(\dfrac{8}{2}\)

Explanation:

• \(\csc^2\theta+\sec^2\theta = \frac{1}{\sin^2\theta}+\frac{1}{\cos^2\theta} = \frac{\cos^2\theta+\sin^2\theta}{\sin^2\theta\cos^2\theta} = \frac{1}{\sin^2\theta\cos^2\theta}\)
  \(= \frac{1}{2/9} = \frac{9}{2}\)

यदि \(b\cos\theta = a\), then \(\csc\theta + \cot\theta = \_\_\_\_\_\_\) होगा।

• \(\sqrt{\dfrac{1}{b+a}}\)
• \(\sqrt{\dfrac{b-a}{b+a}}\)
• \(\sqrt{\dfrac{b+a}{b-a}}\)
• \(\sqrt{\dfrac{1}{b-a}}\)

Explanation:

• \(\cos\theta = \frac{a}{b}\) \(\Rightarrow \sin\theta = \sqrt{1-\frac{a^2}{b^2}} = \frac{\sqrt{b^2-a^2}}{b}\)
  \(\csc\theta+\cot\theta = \frac{1+\cos\theta}{\sin\theta} = \frac{1+\frac{a}{b}}{\frac{\sqrt{b^2-a^2}}{b}} = \frac{b+a}{\sqrt{(b-a)(b+a)}}\)
  \(= \sqrt{\frac{b+a}{b-a}}\)

यदि \(\csc\theta + \cot\theta = p\), तो \(\dfrac{p^2-1}{p^2+1}\) का मान \(\_\_\_\_\_\) है।

• \(\cos\theta\)
• \(\sin\theta\)
• \(\cot\theta\)
• \(\csc\theta\)

Explanation:

• \(\csc\theta+\cot\theta = p \Rightarrow \csc\theta-\cot\theta = \frac{1}{p}\)
  \(\csc\theta = \frac{p^2+1}{2p}\), \(\cot\theta = \frac{p^2-1}{2p}\)
  \(\cos\theta = \frac{\cot\theta}{\csc\theta} = \frac{p^2-1}{p^2+1}\)

यदि \(\sec^2 A + \tan^2 A = 3\) है, तो \(\cot A\) का मान क्या होगा?

• \(\dfrac{1}{\sqrt{3}}\)
• \(0\)
• \(1\)
• \(\sqrt{3}\)

Explanation:

• \(\sec^2 A+\tan^2 A = 3 \Rightarrow 1+\tan^2 A+\tan^2 A = 3 \Rightarrow \tan^2 A = 1\)
  \(\tan A = 1 \Rightarrow \cot A = 1\)

यदि \(\sec\theta + \dfrac{1}{\sec\theta} = 2\) है, तो \(\sec^{55}\theta + \dfrac{1}{\sec^{55}\theta}\) का मान ज्ञात कीजिए।

• \(2\)
• \(0\)
• \(1\)
• \(55\)

Explanation:

• माना \(\sec\theta = s\): \(s+\frac{1}{s} = 2 \Rightarrow (s-1)^2 = 0 \Rightarrow s = 1\)
  \(\sec^{55}\theta+\frac{1}{\sec^{55}\theta} = 1^{55}+\frac{1}{1^{55}} = 2\)

यदि \((\cos q + \sin q) = \dfrac{31}{25}\) है, तो \(\cos^2 q\) का मान क्या होगा?

• \(\dfrac{522}{625}\)
• \(\dfrac{512}{625}\)
• \(\dfrac{513}{625}\)
• \(\dfrac{576}{625}\)

Explanation:

• \((\cos q+\sin q)^2 = \frac{961}{625} \Rightarrow 1+2\sin q\cos q = \frac{961}{625} \Rightarrow 2\sin q\cos q = \frac{336}{625}\)
  \((\cos q-\sin q)^2 = 1-\frac{336}{625} = \frac{289}{625} \Rightarrow \cos q-\sin q = \frac{17}{25}\)
  जोड़ने पर \(2\cos q = \frac{31+17}{25} = \frac{48}{25} \Rightarrow \cos q = \frac{24}{25}\)
  \(\cos^2 q = \frac{576}{625}\)
  नोट: \(\cos q-\sin q = -\frac{17}{25}\) लेने पर \(\cos^2 q = \frac{49}{625}\) भी संभव है, परंतु यह विकल्पों में नहीं है।

यदि \(\sec^2\theta + \tan^2\theta = \dfrac{25}{18}\) है, तो \(\sec^4\theta – \tan^4\theta\) का मान ज्ञात करें:

• \(\dfrac{18}{25}\)
• \(\dfrac{25}{12}\)
• \(\dfrac{28}{9}\)
• \(\dfrac{25}{18}\)

Explanation:

• \(\sec^4\theta-\tan^4\theta = (\sec^2\theta-\tan^2\theta)(\sec^2\theta+\tan^2\theta)\)
  \(= 1\times \frac{25}{18} = \frac{25}{18}\)

यदि \(\sec\theta + \cos\theta = 32\) है, तो \(\sec^2\theta + \cos^2\theta\) का मान ज्ञात कीजिए:

• \(1000\)
• \(1022\)
• \(1024\)
• \(1020\)

Explanation:

• \(\sec^2\theta+\cos^2\theta = (\sec\theta+\cos\theta)^2-2\sec\theta\cos\theta\)
  \(= 32^2-2 = 1024-2 = 1022\)

\(\dfrac{\sin^4\theta + \cos^4\theta}{1 – 2\sin^2\theta\cos^2\theta} = \_\_\_\_\_\_\)

• \(1\)
• \(2\)
• \(-1\)
• \(0\)

Explanation:

• \(\sin^4\theta+\cos^4\theta = (\sin^2\theta+\cos^2\theta)^2-2\sin^2\theta\cos^2\theta = 1-2\sin^2\theta\cos^2\theta\)
  इसलिए व्यंजक \(= \frac{1-2\sin^2\theta\cos^2\theta}{1-2\sin^2\theta\cos^2\theta} = 1\)

निम्नलिखित विकल्पों में से गलत सूत्र का चयन कीजिए।

• \(\sec^2\theta – \tan^2\theta = 1\)
• \(\sin^2\theta + \cos^2\theta = 1\)
• \(\csc^2\theta – \cot^2\theta = 1\)
• \(\sec^2\theta + \cos^2\theta = 1\)

Explanation:

• सही सूत्र: \(\sec^2\theta-\tan^2\theta = 1\), \(\sin^2\theta+\cos^2\theta = 1\), \(\csc^2\theta-\cot^2\theta = 1\)
  जाँच: \(\theta = 0^\circ\) पर \(\sec^2\theta+\cos^2\theta = 1+1 = 2 \neq 1\)
  अतः गलत सूत्र \(\sec^2\theta+\cos^2\theta = 1\) है

\(\left(\dfrac{1}{\sin\theta} + \dfrac{1}{\tan\theta}\right)\left(\dfrac{1}{\sin\theta} – \dfrac{1}{\tan\theta}\right)\) का मान क्या होगा?

• \(0\)
• \(2\)
• \(3\)
• \(1\)

Explanation:

• \(\left(\frac{1}{\sin\theta}+\frac{1}{\tan\theta}\right)\left(\frac{1}{\sin\theta}-\frac{1}{\tan\theta}\right) = (\csc\theta+\cot\theta)(\csc\theta-\cot\theta)\)
  \(= \csc^2\theta-\cot^2\theta = 1\)

निम्नलिखित व्यंजक का मान ज्ञात करें। \(12(\sin^4\theta + \cos^4\theta) + 18(\sin^6\theta + \cos^6\theta) + 78\sin^2\theta\cos^2\theta\)

• \(30\)
• \(40\)
• \(10\)
• \(20\)

Explanation:

• माना \(x = \sin^2\theta\cos^2\theta\)
  \(\sin^4\theta+\cos^4\theta = 1-2x\) और \(\sin^6\theta+\cos^6\theta = 1-3x\)
  व्यंजक \(= 12(1-2x)+18(1-3x)+78x = 12-24x+18-54x+78x = 30\)

यदि \(\dfrac{\sin x – \cos x}{\sin x + \cos x} = \dfrac{2}{5}\) है, तो \(\dfrac{1+\cot^2 x}{1-\cot^2 x}\) का मान ज्ञात कीजिए।

• \(2.25\)
• \(1.45\)
• \(3.75\)
• \(5.25\)

Explanation:

• \(\frac{\sin x-\cos x}{\sin x+\cos x} = \frac{2}{5}\); \(\cos x\) से भाग देने पर \(\frac{\tan x-1}{\tan x+1} = \frac{2}{5}\)
  \(5\tan x-5 = 2\tan x+2 \Rightarrow \tan x = \frac{7}{3} \Rightarrow \cot x = \frac{3}{7}\)
  \(\frac{1+\cot^2 x}{1-\cot^2 x} = \frac{1+\frac{9}{49}}{1-\frac{9}{49}} = \frac{58}{40} = 1.45\)

यदि \(\left(\dfrac{\cos A}{1-\sin A}\right) + \left(\dfrac{\cos A}{1+\sin A}\right) = 4\) है, तो \(A\) का मान क्या होगा? \((0^\circ < \theta < 90^\circ)\)

• \(90^\circ\)
• \(45^\circ\)
• \(60^\circ\)
• \(30^\circ\)

Explanation:

• \(\frac{\cos A}{1-\sin A}+\frac{\cos A}{1+\sin A} = \cos A\cdot\frac{(1+\sin A)+(1-\sin A)}{1-\sin^2 A} = \frac{2\cos A}{\cos^2 A} = 2\sec A\)
  \(2\sec A = 4 \Rightarrow \sec A = 2 \Rightarrow \cos A = \frac{1}{2} \Rightarrow A = 60^\circ\)

यदि \(\cos x + \sin x = \sqrt{2}\cos x\), तो \((\cos x – \sin x)^2 + (\cos x + \sin x)^2\) का मान क्या होगा?

• \(2\)
• \(1\)
• \(0\)
• \(\dfrac{1}{\sqrt{2}}\)

Explanation:

• \((\cos x-\sin x)^2+(\cos x+\sin x)^2 = 2(\cos^2 x+\sin^2 x) = 2\times 1 = 2\)
  (यह मान दी गई शर्त पर निर्भर नहीं करता)

\(\dfrac{\sec^2(90^\circ – \theta) – \cot^2\theta}{2(\sin^2 35^\circ + \sin^2 55^\circ)} = ?\)

• \(0.5\)
• \(2\)
• \(1.5\)
• \(1\)

Explanation:

• \(\sec(90^\circ-\theta) = \csc\theta\), इसलिए अंश \(= \csc^2\theta-\cot^2\theta = 1\)
  \(\sin 55^\circ = \cos 35^\circ \Rightarrow \sin^2 35^\circ+\sin^2 55^\circ = \sin^2 35^\circ+\cos^2 35^\circ = 1\)
  हर \(= 2\times 1 = 2\)
  मान \(= \frac{1}{2} = 0.5\)

निम्नलिखित में कौन-सा विकल्प एक सही त्रिकोणमिति सर्वसमिका नहीं है?

• \(1 + \cot^2\theta = \csc^2\theta\)
• \(1 + \tan^2\theta = \sec^2\theta\)
• \(1 – \sin^2\theta = \cos^2\theta\)
• \(\cos^2\theta – \sin^2\theta = 1\)

Explanation:

• \(\cos^2\theta-\sin^2\theta = \cos 2\theta\), जो सदैव \(1\) के बराबर नहीं होता (उदाहरण \(\theta = 45^\circ\): \(\cos^2\theta-\sin^2\theta = 0 \neq 1\))
  शेष तीनों सर्वसमिकाएँ सही हैं, इसलिए \(\cos^2\theta-\sin^2\theta = 1\) सही सर्वसमिका नहीं है

यदि \(\csc\theta + \cot\theta = m\) है, तो \(\dfrac{m^2-1}{m^2+1}\) का मान ज्ञात कीजिए।

• \(1\)
• \(0\)
• \(\cos\theta\)
• \(-\sin\theta\)

Explanation:

• \(\csc\theta+\cot\theta = m \Rightarrow \csc\theta-\cot\theta = \frac{1}{m}\)
  \(\csc\theta = \frac{m^2+1}{2m}\), \(\cot\theta = \frac{m^2-1}{2m}\)
  \(\frac{m^2-1}{m^2+1} = \frac{\cot\theta}{\csc\theta} = \cos\theta\)

यदि \(\sin\theta + \csc\theta = \sqrt{5}\) है, तो \(\sin^3\theta + \csc^3\theta = \_\_\_\_\) होगा।

• \(0\)
• \(3\sqrt{5}\)
• \(\dfrac{1}{\sqrt{5}}\)
• \(2\sqrt{5}\)

Explanation:

• \(\sin\theta+\csc\theta = \sqrt{5}\)
  घन लेने पर \((\sin\theta+\csc\theta)^3 = \sin^3\theta+\csc^3\theta+3\sin\theta\csc\theta(\sin\theta+\csc\theta)\)
  \(5\sqrt{5} = \sin^3\theta+\csc^3\theta+3\sqrt{5}\)
  \(\sin^3\theta+\csc^3\theta = 5\sqrt{5}-3\sqrt{5} = 2\sqrt{5}\)

\(\sec\theta\sqrt{1 – \sin^2\theta} = ?\)

• \(-1\)
• \(1\)
• \(\infty\)
• \(0\)

Explanation:

• \(\sec\theta\sqrt{1-\sin^2\theta} = \sec\theta\sqrt{\cos^2\theta} = \sec\theta\cos\theta = \frac{1}{\cos\theta}\times \cos\theta = 1\)

यदि \(\cos A + \cos^2 A = 1\) है, तो \(\sin^4 A + \sin^6 A\) का मान है:

• \(2\)
• \(\cos A\)
• \(1\)
• \(\sin A\)

Explanation:

• \(\cos A+\cos^2 A = 1 \Rightarrow \cos A = 1-\cos^2 A = \sin^2 A\)
  \(\sin^4 A = \cos^2 A\) और \(\sin^6 A = \cos^3 A\)
  \(\sin^4 A+\sin^6 A = \cos^2 A+\cos^3 A = \cos A(\cos A+\cos^2 A) = \cos A\times 1 = \cos A\)

यदि \(\cos x + \sec x = \dfrac{7}{2\sqrt{3}}\) हो, तो \(\cos^2 x + \sec^2 x\) का मान ज्ञात कीजिए।

• \(\dfrac{15}{12}\)
• \(\dfrac{10}{12}\)
• \(\dfrac{25}{10}\)
• \(\dfrac{25}{12}\)

Explanation:

• \(\cos x+\sec x = \frac{7}{2\sqrt{3}}\) का वर्ग करने पर
  \(\cos^2 x+\sec^2 x+2\cos x\sec x = \frac{49}{12}\)
  \(\cos^2 x+\sec^2 x = \frac{49}{12}-2 = \frac{25}{12}\)

\(\tan 72^\circ – \tan 27^\circ – \tan 72^\circ\tan 27^\circ\) का मान ज्ञात कीजिए।

• \(-1\)
• \(1\)
• \(-2\)
• \(0\)

Explanation:

• \(\tan(72^\circ-27^\circ) = \tan 45^\circ = 1 = \frac{\tan 72^\circ-\tan 27^\circ}{1+\tan 72^\circ\tan 27^\circ}\)
  \(\Rightarrow \tan 72^\circ-\tan 27^\circ = 1+\tan 72^\circ\tan 27^\circ\)
  \(\Rightarrow \tan 72^\circ-\tan 27^\circ-\tan 72^\circ\tan 27^\circ = 1\)

\(\sin A\sin B = \_\_\_\_\_\_\_\_\)

• \(\dfrac{1}{2}\{\sin(A+B) + \sin(A-B)\}\)
• \(\dfrac{1}{2}\{\sin(A+B) – \sin(A-B)\}\)
• \(\dfrac{1}{2}\{\cos(A+B) – \cos(A-B)\}\)
• \(\dfrac{1}{2}\{\cos(A-B) – \cos(A+B)\}\)

Explanation:

• \(\cos(A-B) = \cos A\cos B+\sin A\sin B\) और \(\cos(A+B) = \cos A\cos B-\sin A\sin B\)
  घटाने पर \(\cos(A-B)-\cos(A+B) = 2\sin A\sin B\)
  \(\sin A\sin B = \frac{1}{2}\{\cos(A-B)-\cos(A+B)\}\)

यदि \(p\sin A – \cos A = 1\), तो \(p^2 – (1 + p^2)\cos A\) का मान ज्ञात कीजिए।

• \(1\)
• \(-1\)
• \(2\)
• \(0\)

Explanation:

• \(p\sin A = 1+\cos A\); वर्ग करने पर \(p^2\sin^2 A = (1+\cos A)^2\)
  \(p^2(1-\cos A)(1+\cos A) = (1+\cos A)^2 \Rightarrow p^2(1-\cos A) = 1+\cos A\)
  \(p^2-p^2\cos A = 1+\cos A \Rightarrow p^2-(1+p^2)\cos A = 1\)

\(A = 60^\circ\) होने पर \(\left(\dfrac{\csc A}{\csc A – 1} + \dfrac{\csc A}{\csc A + 1}\right)\left(\dfrac{\sec A}{\sec A – 1} + \dfrac{\sec A}{\sec A + 1}\right)^{-1}\) का मान क्या होगा:

• \(4\)
• \(3\)
• \(1\)
• \(2\)

Explanation:

• \(A = 60^\circ\): \(\csc^2 A = \frac{4}{3}\), \(\sec^2 A = 4\)
  \(\frac{\csc A}{\csc A-1}+\frac{\csc A}{\csc A+1} = \frac{2\csc^2 A}{\csc^2 A-1} = \frac{2\csc^2 A}{\cot^2 A} = \frac{2}{\cos^2 A} = 2\sec^2 A = 8\)
  \(\frac{\sec A}{\sec A-1}+\frac{\sec A}{\sec A+1} = \frac{2\sec^2 A}{\sec^2 A-1} = \frac{2\sec^2 A}{\tan^2 A} = \frac{2}{\sin^2 A} = 2\csc^2 A = \frac{8}{3}\)
  मान \(= 8\times \left(\frac{8}{3}\right)^{-1} = 8\times \frac{3}{8} = 3\)

यदि \(\sin A + \cos A = \sqrt{3}\), तो \(\tan A + \cot A + 2\sin A\cos A\) का मान ज्ञात कीजिए।

• \(6\)
• \(2\)
• \(1\)
• \(3\)

Explanation:

• \(\sin A+\cos A = \sqrt{3}\) का वर्ग: \(1+2\sin A\cos A = 3 \Rightarrow \sin A\cos A = 1\)
  \(\tan A+\cot A = \frac{\sin^2 A+\cos^2 A}{\sin A\cos A} = \frac{1}{1} = 1\)
  मान \(= 1+2\times 1 = 3\)
  नोट: वास्तविक कोण के लिए \(\sin A\cos A \le \frac{1}{2}\) होता है, फिर भी उत्तर कुंजी इसी बीजगणितीय गणना से मिलती है।

यदि \(\sec\theta + \tan\theta = x\), तो \(\sin\theta\) का मान ज्ञात कीजिए।

• \(\dfrac{x^2+1}{1-x^2}\)
• \(\dfrac{x^2-1}{1+2x^2}\)
• \(\dfrac{x^2-1}{1+x^2}\)
• \(\dfrac{1-x^2}{1+x^2}\)

Explanation:

• \(\sec\theta+\tan\theta = x \Rightarrow \sec\theta-\tan\theta = \frac{1}{x}\)
  \(\sec\theta = \frac{x^2+1}{2x}\), \(\tan\theta = \frac{x^2-1}{2x}\)
  \(\sin\theta = \frac{\tan\theta}{\sec\theta} = \frac{x^2-1}{x^2+1}\)

यदि \(x = a\sec\theta\cos\phi,\ y = b\sec\theta\sin\phi\) और \(z = c\tan\theta\) है, तो \(\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} – \dfrac{z^2}{c^2}\) का मान \(\_\_\_\_\_\_\) होगा:

• \(2\)
• \(1\)
• \(3\)
• \(0\)

Explanation:

• \(\frac{x^2}{a^2} = \sec^2\theta\cos^2\phi\) और \(\frac{y^2}{b^2} = \sec^2\theta\sin^2\phi\)
  \(\frac{x^2}{a^2}+\frac{y^2}{b^2} = \sec^2\theta(\cos^2\phi+\sin^2\phi) = \sec^2\theta\)
  \(\frac{z^2}{c^2} = \tan^2\theta\)
  मान \(= \sec^2\theta-\tan^2\theta = 1\)

यदि \(\sin A – \cos A = \dfrac{\sqrt{3}-\sqrt{2}}{2}\), तो \(\sin A \cdot \cos A\) का मान ज्ञात कीजिए।

• \(\dfrac{2\sqrt{3}-1}{4}\)
• \(\dfrac{2\sqrt{2}-1}{8}\)
• \(\dfrac{2\sqrt{6}-1}{8}\)
• \(\dfrac{3\sqrt{2}-1}{4}\)

Explanation:

• \(\sin A-\cos A = \frac{\sqrt{3}-\sqrt{2}}{2}\) का वर्ग
  \(1-2\sin A\cos A = \frac{3+2-2\sqrt{6}}{4} = \frac{5-2\sqrt{6}}{4}\)
  \(2\sin A\cos A = 1-\frac{5-2\sqrt{6}}{4} = \frac{2\sqrt{6}-1}{4}\)
  \(\sin A\cos A = \frac{2\sqrt{6}-1}{8}\)

यदि \(\cot^4\theta – \cot^2\theta = 1\), तो \(\cos^4\theta + \cos^2\theta\) का मान ज्ञात कीजिए।

• \(2\)
• \(3\)
• \(1\)
• \(\dfrac{3}{2}\)

Explanation:

• माना \(\cot^2\theta = c\): \(c^2-c = 1 \Rightarrow c^2 = c+1\)
  \(\cos^2\theta = \frac{\cot^2\theta}{1+\cot^2\theta} = \frac{c}{1+c}\)
  \(\cos^4\theta+\cos^2\theta = \frac{c^2}{(1+c)^2}+\frac{c}{1+c} = \frac{c^2+c(1+c)}{(1+c)^2} = \frac{2c^2+c}{(1+c)^2}\)
  \(c^2 = c+1\) रखने पर: अंश \(= 2(c+1)+c = 3c+2\); हर \(= c^2+2c+1 = (c+1)+2c+1 = 3c+2\)
  मान \(= 1\)

\((1 + \cot^2\theta)(1 – \cos^2\theta)\) किसके बराबर है?

• \(0\)
• \(1\)
• परिभाषित नहीं
• \(\dfrac{1}{2}\)

Explanation:

• \((1+\cot^2\theta)(1-\cos^2\theta) = \csc^2\theta\times \sin^2\theta\)
  \(= \frac{1}{\sin^2\theta}\times \sin^2\theta = 1\)

\(1 + \dfrac{\tan^2 A}{1 + \sec A} = ?\)

• \(\sec A\)
• \(\csc A\)
• \(\cos a\)
• \(\sin A\)

Explanation:

• \(\tan^2 A = \sec^2 A-1 = (\sec A-1)(\sec A+1)\)
  \(1+\frac{\tan^2 A}{1+\sec A} = 1+\frac{(\sec A-1)(\sec A+1)}{1+\sec A} = 1+\sec A-1 = \sec A\)

\((\csc A – \sin A)^2 + (\sec A – \cos A)^2 – (\cot A – \tan A)^2\) के बराबर है:

• \(1\)
• \(2\)
• \(0\)
• \(-1\)

Explanation:

• \((\csc A-\sin A)^2 = \csc^2 A-2+\sin^2 A\)
  \((\sec A-\cos A)^2 = \sec^2 A-2+\cos^2 A\)
  \((\cot A-\tan A)^2 = \cot^2 A+\tan^2 A-2\)
  व्यंजक \(= (\csc^2 A-\cot^2 A)+(\sec^2 A-\tan^2 A)+(\sin^2 A+\cos^2 A)-2-2+2\)
  \(= 1+1+1-2 = 1\)

यदि \(0 \le \theta \le 90^\circ\), और \(\sec^{107}\theta + \cos^{107}\theta = 2\), तो \((\sec\theta + \cos\theta)\) का मान ज्ञात कीजिए।

• \(2^{-107}\)
• \(1\)
• \(2\)
• \(\dfrac{1}{2}\)

Explanation:

• \(\sec\theta \ge 1\) (\(0\le\theta<90^\circ\)); माना \(x = \sec^{107}\theta \ge 1\), तब \(\cos^{107}\theta = \frac{1}{x}\)
  \(x+\frac{1}{x} = 2 \Rightarrow (x-1)^2 = 0 \Rightarrow x = 1\)
  \(\sec\theta = 1\) और \(\cos\theta = 1\)
  \(\sec\theta+\cos\theta = 1+1 = 2\)

यदि \(\dfrac{1}{\csc\theta + 1} + \dfrac{1}{\csc\theta – 1} = 2\sec\theta\), \(0^\circ < \theta < 90^\circ\) है, तो \(\dfrac{\tan\theta + 2\sec\theta}{\csc\theta}\) का मान है:

• \(\dfrac{2+\sqrt{3}}{2}\)
• \(\dfrac{4+\sqrt{2}}{2}\)
• \(\dfrac{2+\sqrt{2}}{2}\)
• \(\dfrac{4+\sqrt{3}}{2}\)

Explanation:

• \(\frac{1}{\csc\theta+1}+\frac{1}{\csc\theta-1} = \frac{2\csc\theta}{\csc^2\theta-1} = \frac{2\csc\theta}{\cot^2\theta} = \frac{2\sin\theta}{\cos^2\theta}\)
  \(\frac{2\sin\theta}{\cos^2\theta} = \frac{2}{\cos\theta} \Rightarrow \sin\theta = \cos\theta \Rightarrow \theta = 45^\circ\)
  \(\frac{\tan\theta+2\sec\theta}{\csc\theta} = \frac{1+2\sqrt{2}}{\sqrt{2}} = \frac{1}{\sqrt{2}}+2 = \frac{\sqrt{2}+4}{2}\)

यदि \(3 + \cos^2\theta = 3(\cot^2\theta + \sin^2\theta)\), \(0^\circ < \theta < 90^\circ\) है, तो \((\cos\theta + 2\sin\theta)\) का मान ज्ञात कीजिए।

• \(3\sqrt{2}\)
• \(\dfrac{\sqrt{3}+2}{2}\)
• \(\dfrac{2\sqrt{3}+1}{2}\)
• \(\dfrac{3\sqrt{3}+1}{2}\)

Explanation:

• \(\sin^2\theta = s\) लें, तो \(\cos^2\theta = 1-s\), \(\cot^2\theta = \frac{1-s}{s}\)
  \(3+(1-s) = 3\cdot\frac{1-s}{s}+3s \Rightarrow 4s-s^2 = 3-3s+3s^2\)
  \(4s^2-7s+3 = 0 \Rightarrow (4s-3)(s-1) = 0 \Rightarrow s = \frac{3}{4}\) (\(s = 1\) पर \(\theta = 90^\circ\) मान्य नहीं)
  \(\sin\theta = \frac{\sqrt{3}}{2}\), \(\cos\theta = \frac{1}{2}\) (\(\theta = 60^\circ\))
  \(\cos\theta+2\sin\theta = \frac{1}{2}+\sqrt{3} = \frac{2\sqrt{3}+1}{2}\)

यदि \(2\sin\theta + 15\cos^2\theta = 7,\ 0^\circ < \theta < 90^\circ\) है, तो \(\dfrac{3 – \tan\theta}{2 + \tan\theta}\) का मान क्या है?

• \(\dfrac{1}{4}\)
• \(\dfrac{1}{2}\)
• \(\dfrac{5}{8}\)
• \(\dfrac{3}{4}\)

Explanation:

• \(2\sin\theta+15(1-\sin^2\theta) = 7 \Rightarrow 15\sin^2\theta-2\sin\theta-8 = 0\)
  \(\sin\theta = \frac{2+\sqrt{4+480}}{30} = \frac{2+22}{30} = \frac{4}{5}\)
  \(\tan\theta = \frac{4}{3}\)
  \(\frac{3-\tan\theta}{2+\tan\theta} = \frac{3-\frac{4}{3}}{2+\frac{4}{3}} = \frac{5/3}{10/3} = \frac{1}{2}\)

यदि \(\csc\theta = 1.25\), तो \(\dfrac{4\tan\theta – 5\cos\theta + 1}{\sec\theta + 4\cot\theta – 1} = ?\)

• \(2\)
• \(\dfrac{10}{11}\)
• \(\dfrac{9}{10}\)
• \(\dfrac{1}{2}\)

Explanation:

• \(\csc\theta = 1.25 = \frac{5}{4}\) \(\Rightarrow \sin\theta = \frac{4}{5}\), \(\cos\theta = \frac{3}{5}\)
  \(\tan\theta = \frac{4}{3}\), \(\sec\theta = \frac{5}{3}\), \(\cot\theta = \frac{3}{4}\)
  अंश \(= 4\times \frac{4}{3}-5\times \frac{3}{5}+1 = \frac{16}{3}-3+1 = \frac{10}{3}\)
  हर \(= \frac{5}{3}+4\times \frac{3}{4}-1 = \frac{5}{3}+2 = \frac{11}{3}\)
  मान \(= \frac{10}{11}\)

यदि \(\sqrt{2}\sin(60^\circ – a) = 1,\ 0^\circ < a < 90^\circ\) है, तो \(a\) का मान क्या होगा?

• \(30^\circ\)
• \(15^\circ\)
• \(45^\circ\)
• \(45^\circ\)

Explanation:

• \(\sqrt{2}\sin(60^\circ-a) = 1 \Rightarrow \sin(60^\circ-a) = \frac{1}{\sqrt{2}} = \sin 45^\circ\)
  \(60^\circ-a = 45^\circ \Rightarrow a = 15^\circ\)

यदि \(\sin\theta – \cos\theta = 0,\ 0^\circ < \theta < 90^\circ\) है, तो \(\sin^4\theta + \cos^4\theta\) का मान क्या होगा?

• \(\dfrac{1}{3}\)
• \(1\)
• \(\dfrac{1}{4}\)
• \(\dfrac{1}{2}\)

Explanation:

• \(\sin\theta-\cos\theta = 0 \Rightarrow \sin\theta = \cos\theta \Rightarrow \theta = 45^\circ\)
  \(\sin^4 45^\circ+\cos^4 45^\circ = \frac{1}{4}+\frac{1}{4} = \frac{1}{2}\)

यदि \(3(\cot^2\theta – \cos^2\theta) = 1 – \sin^2\theta,\ 0^\circ < \theta < 90^\circ\) है, तो \(\theta\) का मान बताइए।

• \(30^\circ\)
• \(60^\circ\)
• \(45^\circ\)
• \(15^\circ\)

Explanation:

• \(\cot^2\theta-\cos^2\theta = \cos^2\theta\left(\frac{1}{\sin^2\theta}-1\right) = \frac{\cos^4\theta}{\sin^2\theta}\)
  \(1-\sin^2\theta = \cos^2\theta\)
  \(\frac{3\cos^4\theta}{\sin^2\theta} = \cos^2\theta \Rightarrow 3\cos^2\theta = \sin^2\theta \Rightarrow \tan^2\theta = 3\)
  \(\tan\theta = \sqrt{3} \Rightarrow \theta = 60^\circ\)

यदि \(\cos^2\theta – \sin^2\theta = \tan^2\phi\) है, तो निम्नलिखित में से कौन-सा विकल्प सही है?

• \(\cos\theta\cos\phi = \sqrt{2}\)
• \(\cos\theta\cos\phi = 1\)
• \(\cos^2\phi – \sin^2\phi = \tan^2\theta\)
• \(\cos^2\phi – \cos\phi = \cot^2\theta\)

Explanation:

• माना \(\cos^2\theta-\sin^2\theta = c = \tan^2\phi\)
  \(\cos^2\phi = \frac{1}{1+\tan^2\phi} = \frac{1}{1+c}\), \(\sin^2\phi = \frac{c}{1+c}\)
  \(\cos^2\phi-\sin^2\phi = \frac{1-c}{1+c}\)
  \(1-c = 1-\cos^2\theta+\sin^2\theta = 2\sin^2\theta\) और \(1+c = 2\cos^2\theta\)
  \(\cos^2\phi-\sin^2\phi = \frac{2\sin^2\theta}{2\cos^2\theta} = \tan^2\theta\)

\(\dfrac{\sin^2 52^\circ + 2 + \sin^2 38^\circ}{4\cos^2 43^\circ – 5 + 4\cos^2 47^\circ} = ?\)

• \(3\)
• \(\dfrac{1}{3}\)
• \(-\dfrac{1}{3}\)
• \(-3\)

Explanation:

• \(\sin 38^\circ = \cos 52^\circ \Rightarrow \sin^2 52^\circ+\sin^2 38^\circ = \sin^2 52^\circ+\cos^2 52^\circ = 1\)
  अंश \(= 1+2 = 3\)
  \(\cos 47^\circ = \sin 43^\circ \Rightarrow \cos^2 43^\circ+\cos^2 47^\circ = 1\)
  हर \(= 4(1)-5 = -1\)
  मान \(= \frac{3}{-1} = -3\)

यदि \(4(\csc^2 57^\circ – \tan^2 33^\circ) – \cos 90^\circ + y\tan^2 66^\circ\tan^2 24^\circ = \dfrac{y}{2}\) है, तो \(y\) का मान ज्ञात करें।

• \(4\)
• \(-4\)
• \(8\)
• \(-8\)

Explanation:

• \(\csc 57^\circ = \sec 33^\circ \Rightarrow \csc^2 57^\circ-\tan^2 33^\circ = \sec^2 33^\circ-\tan^2 33^\circ = 1\)
  \(\cos 90^\circ = 0\); \(\tan 66^\circ\tan 24^\circ = \tan 66^\circ\cot 66^\circ = 1\), इसलिए \(\tan^2 66^\circ\tan^2 24^\circ = 1\)
  \(4(1)-0+y\times 1 = \frac{y}{2} \Rightarrow \frac{y}{2} = -4 \Rightarrow y = -8\)

\(\theta\) का मान ज्ञात करें: \(\cos^2\theta – \sin^2\theta = \dfrac{1}{2}\) जहां \(0 < \theta < 90^\circ\) है।

• \(45^\circ\)
• \(60^\circ\)
• \(30^\circ\)
• \(40^\circ\)

Explanation:

• \(\cos^2\theta-\sin^2\theta = \cos 2\theta = \frac{1}{2} = \cos 60^\circ\)
  \(2\theta = 60^\circ \Rightarrow \theta = 30^\circ\)

यदि \(\cot\theta = \dfrac{1}{\sqrt{3}},\ 0^\circ < \theta < 90^\circ\), तो \(\dfrac{2 – \sin^2\theta}{1 – \cos^2\theta} + (\csc^2\theta – \sec\theta)\) का मान ज्ञात करें।

• \(2\)
• \(5\)
• \(0\)
• \(1\)

Explanation:

• \(\cot\theta = \frac{1}{\sqrt{3}} \Rightarrow \theta = 60^\circ\); \(\sin^2\theta = \frac{3}{4}\), \(\cos^2\theta = \frac{1}{4}\), \(\csc^2\theta = \frac{4}{3}\), \(\sec\theta = 2\)
  \(\frac{2-\sin^2\theta}{1-\cos^2\theta} = \frac{2-\frac{3}{4}}{1-\frac{1}{4}} = \frac{5/4}{3/4} = \frac{5}{3}\)
  \(\csc^2\theta-\sec\theta = \frac{4}{3}-2 = -\frac{2}{3}\)
  कुल \(= \frac{5}{3}-\frac{2}{3} = 1\)

\(23\csc^2 A – 23\cot^2 A = ?\)

• \(23\)
• \(1\)
• \(27\)
• \(0\)

Explanation:

• \(23\csc^2 A-23\cot^2 A = 23(\csc^2 A-\cot^2 A) = 23\times 1 = 23\)

\((\csc A + \cot A)(1 – \cos A) = ?\)

• \(\cos A\)
• \(\tan A\)
• \(\cot A\)
• \(\sin A\)

Explanation:

• \((\csc A+\cot A)(1-\cos A) = \frac{1+\cos A}{\sin A}\times (1-\cos A)\)
  \(= \frac{1-\cos^2 A}{\sin A} = \frac{\sin^2 A}{\sin A} = \sin A\)

\(\dfrac{2\cos^3\theta – \cos\theta}{\sin\theta – 2\sin^3\theta} = ?\)

• \(\sec\theta\)
• \(\sin\theta\)
• \(\cot\theta\)
• \(\tan\theta\)

Explanation:

• अंश \(= \cos\theta(2\cos^2\theta-1) = \cos\theta\cos 2\theta\)
  हर \(= \sin\theta(1-2\sin^2\theta) = \sin\theta\cos 2\theta\)
  मान \(= \frac{\cos\theta}{\sin\theta} = \cot\theta\)

यदि \(1 + \sin^2\theta – 3\sin\theta\cos\theta = 0\), तो \(\cot\theta\) का मान क्या होगा?

• \(2\)
• \(1/3\)
• \(1/2\)
• \(0\)

Explanation:

• \(\sin^2\theta\) से भाग देने पर: \(\csc^2\theta+1-3\cot\theta = 0\)
  \(1+\cot^2\theta+1-3\cot\theta = 0 \Rightarrow \cot^2\theta-3\cot\theta+2 = 0\)
  \((\cot\theta-1)(\cot\theta-2) = 0 \Rightarrow \cot\theta = 1\) या \(2\)
  विकल्पों में केवल \(2\) है, इसलिए \(\cot\theta = 2\)
  नोट: \(\cot\theta = 1\) भी समीकरण को संतुष्ट करता है, परंतु यह विकल्पों में नहीं है।

\(\dfrac{\sin A}{\cot A + \csc A} – \dfrac{\sin A}{\cot A – \csc A} + 1 = ?\)

• \(1/2\)
• \(3\)
• \(0\)
• \(2\)

Explanation:

• \(\frac{\sin A}{\cot A+\csc A}-\frac{\sin A}{\cot A-\csc A} = \sin A\cdot\frac{(\cot A-\csc A)-(\cot A+\csc A)}{\cot^2 A-\csc^2 A}\)
  \(= \sin A\cdot\frac{-2\csc A}{-1} = 2\sin A\csc A = 2\)
  मान \(= 2+1 = 3\)

यदि \(\sin^2\theta = \cos^3\theta\) है, तो \((\cot^2\theta – \cot^6\theta)\) ज्ञात कीजिए?

• \(-1\)
• \(0\)
• \(2\)
• \(1\)

Explanation:

• \(\sin^2\theta = \cos^3\theta \Rightarrow 1-\cos^2\theta = \cos^3\theta\)
  \(\cot^2\theta = \frac{\cos^2\theta}{\sin^2\theta} = \frac{\cos^2\theta}{\cos^3\theta} = \frac{1}{\cos\theta}\)
  माना \(k = \cot^2\theta = \frac{1}{\cos\theta}\); \(1-\cos^2\theta = \cos^3\theta\) को \(\cos^3\theta\) से भाग देने पर \(k^3-k = 1\)
  \(\cot^2\theta-\cot^6\theta = k-k^3 = -1\)

यदि \(\cos\theta – \sin\theta = \sqrt{2}\sin\theta\) तो \((\cos\theta + \sin\theta)\) है:

• \(-\sqrt{2}\cos\theta\)
• \(\sqrt{2}\cos\theta\)
• \(\sqrt{2}\tan\theta\)
• \(-\sqrt{2}\sin\theta\)

Explanation:

• \(\cos\theta-\sin\theta = \sqrt{2}\sin\theta \Rightarrow \cos\theta = (\sqrt{2}+1)\sin\theta\)
  \(\cos\theta+\sin\theta = (\sqrt{2}+1)\sin\theta+\sin\theta = (\sqrt{2}+2)\sin\theta = \sqrt{2}(1+\sqrt{2})\sin\theta\)
  \(= \sqrt{2}\cos\theta\)

\((\tan^2\theta + \tan^4\theta) = ?\)

• \(\cot^2\theta – \tan^2\theta\)
• \(\sec^4\theta – \sec^2\theta\)
• \(\sec^2\theta – \sec^4\theta\)
• \(\sec^2\theta – \sec^2\theta\)

Explanation:

• \(\tan^2\theta+\tan^4\theta = \tan^2\theta(1+\tan^2\theta) = \tan^2\theta\sec^2\theta\)
  \(= (\sec^2\theta-1)\sec^2\theta = \sec^4\theta-\sec^2\theta\)

यदि \(\sin\theta + \cos\theta = \dfrac{\sqrt{11}}{3}\) है, तो \((\cos\theta – \sin\theta)\) का मान है:

• \(\dfrac{\sqrt{5}}{3}\)
• \(\dfrac{7}{3}\)
• \(\dfrac{5}{3}\)
• \(\dfrac{\sqrt{7}}{3}\)

Explanation:

• \((\sin\theta+\cos\theta)^2 = \frac{11}{9} \Rightarrow 1+2\sin\theta\cos\theta = \frac{11}{9} \Rightarrow 2\sin\theta\cos\theta = \frac{2}{9}\)
  \((\cos\theta-\sin\theta)^2 = 1-2\sin\theta\cos\theta = 1-\frac{2}{9} = \frac{7}{9}\)
  \(\cos\theta-\sin\theta = \frac{\sqrt{7}}{3}\)

यदि \(\sec A – \tan A = p\) है, तो \(\sec A\) का मान ज्ञात कीजिए।

• \(\dfrac{p^2-1}{p^2+1}\)
• \(\dfrac{p^2+1}{p^2-1}\)
• \(\dfrac{p^2+1}{p}\)
• \(\dfrac{p^2+1}{2p}\)

Explanation:

• \(\sec A-\tan A = p\) और \(\sec^2 A-\tan^2 A = 1 \Rightarrow \sec A+\tan A = \frac{1}{p}\)
  जोड़ने पर \(2\sec A = p+\frac{1}{p} = \frac{p^2+1}{p}\)
  \(\sec A = \frac{p^2+1}{2p}\)

यदि \(\csc A + \cot A = 3,\ 0^\circ \le A \le 90^\circ\), है, तो \(\cos A\) का मान ज्ञात कीजिए।

• \(\dfrac{3}{4}\)
• \(\dfrac{2}{5}\)
• \(\dfrac{3}{5}\)
• \(\dfrac{4}{5}\)

Explanation:

• \(\csc A+\cot A = 3 \Rightarrow \csc A-\cot A = \frac{1}{3}\)
  जोड़ने पर \(2\csc A = \frac{10}{3} \Rightarrow \csc A = \frac{5}{3}\); घटाने पर \(2\cot A = \frac{8}{3} \Rightarrow \cot A = \frac{4}{3}\)
  \(\cos A = \frac{\cot A}{\csc A} = \frac{4/3}{5/3} = \frac{4}{5}\)

\(\dfrac{\sin^2 39^\circ + \sin^2(90^\circ – 39^\circ)}{\cos^2 35^\circ + \cos^2(90^\circ – 35^\circ)} + 3\tan 25^\circ\tan 65^\circ = ?\)

• \(2\)
• \(4\)
• \(3\)
• \(1\)

Explanation:

• \(\sin(90^\circ-39^\circ) = \cos 39^\circ \Rightarrow \sin^2 39^\circ+\cos^2 39^\circ = 1\)
  \(\cos(90^\circ-35^\circ) = \sin 35^\circ \Rightarrow \cos^2 35^\circ+\sin^2 35^\circ = 1\)
  \(\tan 65^\circ = \cot 25^\circ \Rightarrow \tan 25^\circ\tan 65^\circ = 1\)
  मान \(= \frac{1}{1}+3\times 1 = 4\)

यदि \(\sin\theta + \cos\theta = \sqrt{2}\cos\theta\), तो \(\dfrac{\sin\theta – \cos\theta}{\sin\theta}\) ज्ञात कीजिए।

• \(-\sqrt{2}\)
• \(-1\)
• \(1\)
• \(\sqrt{2}\)

Explanation:

• \(\sin\theta+\cos\theta = \sqrt{2}\cos\theta \Rightarrow \sin\theta = (\sqrt{2}-1)\cos\theta \Rightarrow \cot\theta = \frac{1}{\sqrt{2}-1} = \sqrt{2}+1\)
  \(\frac{\sin\theta-\cos\theta}{\sin\theta} = 1-\cot\theta = 1-(\sqrt{2}+1) = -\sqrt{2}\)

\((1 + \tan^2 A) + \left(1 + \dfrac{1}{\tan^2 A}\right) = ?\)

• \(\dfrac{1}{\sin^2 A – \sin^4 A}\)
• \(\dfrac{1}{\sin^2 A – \cos^4 A}\)
• \(\dfrac{1}{\cos^2 A – \sin^4 A}\)
• \(\dfrac{1}{\cot^2 A – \tan^4 A}\)

Explanation:

• \((1+\tan^2 A)+\left(1+\frac{1}{\tan^2 A}\right) = \sec^2 A+\csc^2 A = \frac{1}{\cos^2 A}+\frac{1}{\sin^2 A} = \frac{1}{\sin^2 A\cos^2 A}\)
  \(\sin^2 A\cos^2 A = \sin^2 A(1-\sin^2 A) = \sin^2 A-\sin^4 A\)
  अतः मान \(= \frac{1}{\sin^2 A-\sin^4 A}\)

यदि \(\tan(t) + \cot(t) = 1\), है तो व्यंजक \(\dfrac{1}{\sin(t) + \cos(t)}\) का मान \(\_\_\_\_\_\) है।

• \(\dfrac{1}{\sqrt{3}}\)
• \(\dfrac{2}{\sqrt{3}}\)
• \(\dfrac{\sqrt{3}}{2}\)
• \(\sqrt{3}\)

Explanation:

• \(\tan t+\cot t = \frac{1}{\sin t\cos t} = 1 \Rightarrow \sin t\cos t = 1\)
  \((\sin t+\cos t)^2 = 1+2\sin t\cos t = 1+2 = 3 \Rightarrow \sin t+\cos t = \sqrt{3}\)
  \(\frac{1}{\sin t+\cos t} = \frac{1}{\sqrt{3}}\)
  नोट: वास्तविक कोण के लिए \(\sin t\cos t \le \frac{1}{2}\) होता है, फिर भी उत्तर कुंजी इसी बीजगणितीय गणना से मिलती है।

यदि \((\sin x + \sin y) = a\) और \((\cos x + \cos y) = b\), तो \((\sin x\sin y + \cos x\cos y)\) का मान ज्ञात कीजिए।

• \(\dfrac{a^2+b^2-1}{2}\)
• \(\dfrac{a^2+b^2-2}{2}\)
• \(\dfrac{a^2+b^2-2}{4}\)
• \(\dfrac{a^2+b^2+2}{2}\)

Explanation:

• \(a^2+b^2 = (\sin x+\sin y)^2+(\cos x+\cos y)^2\)
  \(= (\sin^2 x+\cos^2 x)+(\sin^2 y+\cos^2 y)+2(\sin x\sin y+\cos x\cos y) = 2+2(\sin x\sin y+\cos x\cos y)\)
  \(\sin x\sin y+\cos x\cos y = \frac{a^2+b^2-2}{2}\)
  नोट: उत्तर कुंजी में \(\frac{a^2+b^2-2}{4}\) चिह्नित है, परंतु सही गणना \(\frac{a^2+b^2-2}{2}\) आती है (जो दूसरा विकल्प है)।

\(\dfrac{\cos 13^\circ + \sin 13^\circ}{\cos 13^\circ – \sin 13^\circ} = ?\)

• \(\cot 58^\circ\)
• \(\tan 32^\circ\)
• \(\tan 58^\circ\)
• \(\cos 26^\circ\)

Explanation:

• अंश-हर को \(\cos 13^\circ\) से भाग देने पर \(\frac{1+\tan 13^\circ}{1-\tan 13^\circ} = \frac{\tan 45^\circ+\tan 13^\circ}{1-\tan 45^\circ\tan 13^\circ}\)
  \(= \tan(45^\circ+13^\circ) = \tan 58^\circ\)

यदि \(4\sin^2\theta = 3(1 + \cos\theta),\ 0^\circ < \theta < 90^\circ\), तो \(\sqrt{15}\tan\theta + \dfrac{4}{\sqrt{15}}\sin\theta + 2\sec\theta\) का मान क्या होगा?

• \(24\)
• \(4\sqrt{15}\)
• \(\dfrac{8}{\sqrt{15}}\)
• \(8\)

Explanation:

• \(4\sin^2\theta = 3(1+\cos\theta) \Rightarrow 4(1-\cos\theta)(1+\cos\theta) = 3(1+\cos\theta)\)
  \(4(1-\cos\theta) = 3 \Rightarrow \cos\theta = \frac{1}{4}\)
  \(\sin\theta = \frac{\sqrt{15}}{4}\), \(\tan\theta = \sqrt{15}\), \(\sec\theta = 4\)
  मान \(= \sqrt{15}\times \sqrt{15}+\frac{4}{\sqrt{15}}\times \frac{\sqrt{15}}{4}+2\times 4 = 15+1+8 = 24\)

\((\csc\theta – \sin\theta)(\sec\theta – \cos\theta)(\tan\theta + \cot\theta)\) का मान है:

• \(-3\)
• \(3\)
• \(\sec\theta\csc\theta\)
• \(1\)

Explanation:

• \(\csc\theta-\sin\theta = \frac{\cos^2\theta}{\sin\theta}\), \(\sec\theta-\cos\theta = \frac{\sin^2\theta}{\cos\theta}\)
  \(\tan\theta+\cot\theta = \frac{1}{\sin\theta\cos\theta}\)
  गुणनफल \(= \frac{\cos^2\theta}{\sin\theta}\times \frac{\sin^2\theta}{\cos\theta}\times \frac{1}{\sin\theta\cos\theta} = 1\)

यदि \(\sin x + \cos x = \sqrt{2}\sin x\), तो \(\sin x – \cos x\) का मान है:

• \(\sqrt{2}\sin x\)
• \(2\sin x\)
• \(-\sqrt{2}\cos x\)
• \(\sqrt{2}\cos x\)

Explanation:

• \(\sin x+\cos x = \sqrt{2}\sin x \Rightarrow \cos x = (\sqrt{2}-1)\sin x\)
  \(\sin x-\cos x = \sin x-(\sqrt{2}-1)\sin x = (2-\sqrt{2})\sin x = \sqrt{2}(\sqrt{2}-1)\sin x\)
  \(= \sqrt{2}\cos x\)

\(\left(\dfrac{2\cos^3\theta – \cos\theta}{\sin\theta – 2\sin^3\theta}\right)^2 + 1,\ \theta \ne 45^\circ\) का मान है:

• \(\sin^2\theta\)
• \(\sec^2\theta\)
• \(\csc^2\theta\)
• \(\cot^2\theta\)

Explanation:

• \(\frac{2\cos^3\theta-\cos\theta}{\sin\theta-2\sin^3\theta} = \frac{\cos\theta(2\cos^2\theta-1)}{\sin\theta(1-2\sin^2\theta)} = \frac{\cos\theta\cos 2\theta}{\sin\theta\cos 2\theta} = \cot\theta\)
  (\(1-2\sin^2\theta = 2\cos^2\theta-1 = \cos 2\theta\))
  \(\cot^2\theta+1 = \csc^2\theta\)

यदि \(\sin^4\theta + \cos^4\theta = 2\sin^2\theta\cos^2\theta,\ 0^\circ < \theta < 90^\circ\), तो \(\cot\theta\) का मान ज्ञात कीजिए।

• \(\dfrac{1}{2}\)
• \(\dfrac{1}{\sqrt{2}}\)
• \(\dfrac{1}{3}\)
• \(1\)

Explanation:

• \(\sin^4\theta+\cos^4\theta-2\sin^2\theta\cos^2\theta = 0 \Rightarrow (\sin^2\theta-\cos^2\theta)^2 = 0\)
  \(\sin^2\theta = \cos^2\theta \Rightarrow \tan\theta = 1 \Rightarrow \cot\theta = 1\)

\((\tan\theta + \sec\theta + 1)(\cot\theta – \csc\theta + 1) = ?\)

• \(2\)
• \(1\)
• \(0\)
• \(-1\)

Explanation:

• \(\tan\theta+\sec\theta+1 = \frac{\sin\theta+1+\cos\theta}{\cos\theta}\)
  \(\cot\theta-\csc\theta+1 = \frac{\cos\theta-1+\sin\theta}{\sin\theta}\)
  गुणनफल \(= \frac{(\sin\theta+\cos\theta)^2-1}{\sin\theta\cos\theta} = \frac{2\sin\theta\cos\theta}{\sin\theta\cos\theta} = 2\)

\(\dfrac{\cot\theta}{(1 – \sin\theta)(\sec\theta + \tan\theta)} = ?\)

• \(\csc\theta\)
• \(\sin\theta\)
• \(\sec\theta\)
• \(1\)

Explanation:

• \((1-\sin\theta)(\sec\theta+\tan\theta) = (1-\sin\theta)\cdot\frac{1+\sin\theta}{\cos\theta} = \frac{\cos^2\theta}{\cos\theta} = \cos\theta\)
  \(\frac{\cot\theta}{\cos\theta} = \frac{\cos\theta}{\sin\theta\cos\theta} = \frac{1}{\sin\theta} = \csc\theta\)

यदि \(\cos^2\theta – \sin^2\theta – 3\cos\theta + 2 = 0,\ 0^\circ < \theta < 90^\circ\), तो \(4\csc\theta + \cot\theta\) का मान कितना है?

• \(3\sqrt{3}\)
• \(4\)
• \(4\sqrt{3}\)
• \(3\)

Explanation:

• \(\cos^2\theta-\sin^2\theta-3\cos\theta+2 = 0 \Rightarrow 2\cos^2\theta-1-3\cos\theta+2 = 0\)
  \(2\cos^2\theta-3\cos\theta+1 = 0 \Rightarrow (2\cos\theta-1)(\cos\theta-1) = 0\)
  \(\cos\theta = \frac{1}{2}\) (\(\cos\theta = 1\) पर \(\theta = 0^\circ\) मान्य नहीं) \(\Rightarrow \theta = 60^\circ\)
  \(4\csc 60^\circ+\cot 60^\circ = \frac{8}{\sqrt{3}}+\frac{1}{\sqrt{3}} = \frac{9}{\sqrt{3}} = 3\sqrt{3}\)

यदि \(\dfrac{1}{\sec\theta – \tan\theta} – \dfrac{1}{\cos\theta} = \sec\theta \times k,\ 0^\circ < \theta < 90^\circ\) तो \(k\) बराबर है:

• \(\csc\theta\)
• \(\tan\theta\)
• \(\sin\theta\)
• \(\cot\theta\)

Explanation:

• \(\frac{1}{\sec\theta-\tan\theta} = \sec\theta+\tan\theta\) (क्योंकि \(\sec^2\theta-\tan^2\theta = 1\))
  \(\sec\theta+\tan\theta-\sec\theta = \tan\theta = k\sec\theta\) (\(\frac{1}{\cos\theta} = \sec\theta\))
  \(k = \frac{\tan\theta}{\sec\theta} = \sin\theta\)

यदि \(3\sin\theta = 2\cos^2\theta,\ 0^\circ < \theta < 90^\circ\) है तो \((\tan\theta + \cos\theta + \sin\theta)\) का मान है:

• \(\dfrac{5\sqrt{3}}{3}\)
• \(\dfrac{5\sqrt{3}}{6}\)
• \(\dfrac{3+5\sqrt{3}}{6}\)
• \(\dfrac{3+5\sqrt{3}}{3}\)

Explanation:

• \(3\sin\theta = 2(1-\sin^2\theta) \Rightarrow 2\sin^2\theta+3\sin\theta-2 = 0 \Rightarrow (2\sin\theta-1)(\sin\theta+2) = 0\)
  \(\sin\theta = \frac{1}{2} \Rightarrow \theta = 30^\circ\)
  \(\tan\theta+\cos\theta+\sin\theta = \frac{1}{\sqrt{3}}+\frac{\sqrt{3}}{2}+\frac{1}{2} = \frac{2\sqrt{3}+3\sqrt{3}+3}{6} = \frac{3+5\sqrt{3}}{6}\)

\(\left(\dfrac{\sin\theta}{1+\cos\theta} + \dfrac{1+\cos\theta}{\sin\theta}\right)\left(\dfrac{1}{\tan\theta + \cot\theta}\right) = ?\)

• \(\cos\theta\)
• \(2\sin\theta\)
• \(\sin\theta\)
• \(2\cos\theta\)

Explanation:

• \(\frac{\sin\theta}{1+\cos\theta}+\frac{1+\cos\theta}{\sin\theta} = \frac{\sin^2\theta+(1+\cos\theta)^2}{\sin\theta(1+\cos\theta)} = \frac{2+2\cos\theta}{\sin\theta(1+\cos\theta)} = \frac{2}{\sin\theta}\)
  \(\frac{1}{\tan\theta+\cot\theta} = \sin\theta\cos\theta\)
  गुणनफल \(= \frac{2}{\sin\theta}\times \sin\theta\cos\theta = 2\cos\theta\)

यदि \(\dfrac{(\sin\theta – \csc\theta)(\cos\theta – \sec\theta)}{\tan^2\theta – \sin^2\theta} = r^3\) है, तो \(r = \)

• \(\sin\theta\cos\theta\)
• \(\tan\theta\)
• \(\cot\theta\)
• \(\csc\theta\sin\theta\)

Explanation:

• \((\sin\theta-\csc\theta)(\cos\theta-\sec\theta) = \left(-\frac{\cos^2\theta}{\sin\theta}\right)\left(-\frac{\sin^2\theta}{\cos\theta}\right) = \sin\theta\cos\theta\)
  \(\tan^2\theta-\sin^2\theta = \sin^2\theta\left(\frac{1}{\cos^2\theta}-1\right) = \frac{\sin^4\theta}{\cos^2\theta}\)
  \(r^3 = \frac{\sin\theta\cos\theta\cdot\cos^2\theta}{\sin^4\theta} = \frac{\cos^3\theta}{\sin^3\theta} = \cot^3\theta \Rightarrow r = \cot\theta\)

यदि \(\left(\dfrac{1}{1+\csc\theta} – \dfrac{1}{1-\csc\theta}\right)\cos\theta = 2,\ 0^\circ < \theta < 90^\circ\) है, तो \(\sin^2\theta + \cot^2\theta + \sec^2\theta\) का मान है:

• \(1\)
• \(2\tfrac{1}{2}\)
• \(3\tfrac{1}{2}\)
• \(2\)

Explanation:

• \(\frac{1}{1+\csc\theta}-\frac{1}{1-\csc\theta} = \frac{(1-\csc\theta)-(1+\csc\theta)}{1-\csc^2\theta} = \frac{-2\csc\theta}{-\cot^2\theta} = \frac{2\csc\theta}{\cot^2\theta}\)
  \(\times\cos\theta\): \(\frac{2\csc\theta\cos\theta}{\cot^2\theta} = \frac{2\cdot\frac{\cos\theta}{\sin\theta}}{\frac{\cos^2\theta}{\sin^2\theta}} = 2\tan\theta\)
  \(2\tan\theta = 2 \Rightarrow \tan\theta = 1 \Rightarrow \theta = 45^\circ\)
  \(\sin^2\theta+\cot^2\theta+\sec^2\theta = \frac{1}{2}+1+2 = 3\frac{1}{2}\)

यदि \(2(\csc^2 39^\circ – \tan^2 51^\circ) – \dfrac{2}{3}\sin 90^\circ – \tan^2 56^\circ \cdot y \cdot \tan^2 34^\circ = \dfrac{y}{3}\) है, तो \(y\) का मान है?

• \(1\)
• \(\dfrac{2}{3}\)
• \(-\dfrac{2}{3}\)
• \(-1\)

Explanation:

• \(\csc 39^\circ = \sec 51^\circ \Rightarrow \csc^2 39^\circ-\tan^2 51^\circ = \sec^2 51^\circ-\tan^2 51^\circ = 1\)
  \(\sin 90^\circ = 1\); \(\tan 56^\circ\tan 34^\circ = 1 \Rightarrow \tan^2 56^\circ\tan^2 34^\circ = 1\)
  \(2(1)-\frac{2}{3}-y = \frac{y}{3} \Rightarrow \frac{4}{3} = \frac{4y}{3} \Rightarrow y = 1\)

अगर \(\theta\) एक न्यून कोण है और \(3(\csc^2\theta + \cot^2\theta) = 5\) दिया गया है तो \(\theta\) किसके बराबर होगा?

• \(45^\circ\)
• \(60^\circ\)
• \(0^\circ\)
• \(30^\circ\)

Explanation:

• \(3(\csc^2\theta+\cot^2\theta) = 5 \Rightarrow \csc^2\theta+\cot^2\theta = \frac{5}{3}\)
  \(1+\cot^2\theta+\cot^2\theta = \frac{5}{3} \Rightarrow 2\cot^2\theta = \frac{2}{3} \Rightarrow \cot^2\theta = \frac{1}{3}\)
  \(\cot\theta = \frac{1}{\sqrt{3}} \Rightarrow \theta = 60^\circ\)

यदि \(117\cos^2 A + 129\sin^2 A = 120\) है और \(170\cos^2 B + 158\sin^2 B = 161\) है, तो \(\csc^2 A\sec^2 B\) का मान ज्ञात करें।

• \(1\)
• \(9\)
• \(4\)
• \(16\)

Explanation:

• \(117\cos^2 A+129\sin^2 A = 120 \Rightarrow 117(\cos^2 A+\sin^2 A)+12\sin^2 A = 120\)
  \(117+12\sin^2 A = 120 \Rightarrow \sin^2 A = \frac{1}{4} \Rightarrow \csc^2 A = 4\)
  \(170\cos^2 B+158\sin^2 B = 161 \Rightarrow 158+12\cos^2 B = 161 \Rightarrow \cos^2 B = \frac{1}{4} \Rightarrow \sec^2 B = 4\)
  \(\csc^2 A\sec^2 B = 4\times 4 = 16\)

अगर \(\sec A = \dfrac{\sqrt{11}}{3}\) है, तो \(\dfrac{\csc^2 A + \tan^2 A}{\sin^2 A + \cot^2 A}\) का मान ज्ञात कीजिए।

• \(\dfrac{9}{4}\)
• \(\dfrac{4}{9}\)
• \(\dfrac{11}{9}\)
• \(\dfrac{2}{11}\)

Explanation:

• \(\sec A = \frac{\sqrt{11}}{3} \Rightarrow \cos^2 A = \frac{9}{11},\ \sin^2 A = \frac{2}{11}\)
  \(\csc^2 A = \frac{11}{2},\ \tan^2 A = \sec^2 A-1 = \frac{2}{9},\ \cot^2 A = \frac{9}{2}\)
  अंश \(= \frac{11}{2}+\frac{2}{9} = \frac{103}{18}\); हर \(= \frac{2}{11}+\frac{9}{2} = \frac{103}{22}\)
  मान \(= \frac{103}{18}\times \frac{22}{103} = \frac{11}{9}\)

\(\dfrac{\cot x}{1 + \csc x} + \dfrac{1 + \csc x}{\cot x} = ?\)

• \(2\sec x\)
• \(2\cos x\)
• \(2\csc x\)
• \(2\sin x\)

Explanation:

• \(\frac{\cot x}{1+\csc x}+\frac{1+\csc x}{\cot x} = \frac{\cot^2 x+(1+\csc x)^2}{\cot x(1+\csc x)}\)
  \(\cot^2 x = \csc^2 x-1\), अतः अंश \(= \csc^2 x-1+1+2\csc x+\csc^2 x = 2\csc x(\csc x+1)\)
  मान \(= \frac{2\csc x(1+\csc x)}{\cot x(1+\csc x)} = \frac{2\csc x}{\cot x} = \frac{2}{\cos x} = 2\sec x\)

यदि \(A\) और \(B\) न्यून कोण हैं और \(\sec A = 3;\ \cot B = 4\) है, तो \(\dfrac{\csc^2 A + \sin^2 B}{\cot^2 A + \sec^2 B}\) का मान ज्ञात कीजिए।

• \(\dfrac{25}{261}\)
• \(\dfrac{322}{323}\)
• \(2\)
• \(\dfrac{1}{261}\)

Explanation:

• \(\sec A = 3 \Rightarrow \cos A = \frac{1}{3},\ \sin^2 A = \frac{8}{9},\ \csc^2 A = \frac{9}{8},\ \cot^2 A = \frac{1}{8}\)
  \(\cot B = 4 \Rightarrow \sin^2 B = \frac{1}{17},\ \sec^2 B = 1+\tan^2 B = 1+\frac{1}{16} = \frac{17}{16}\)
  अंश \(= \frac{9}{8}+\frac{1}{17} = \frac{161}{136}\); हर \(= \frac{1}{8}+\frac{17}{16} = \frac{19}{16}\)
  मान \(= \frac{161}{136}\times \frac{16}{19} = \frac{322}{323}\)

यदि \(\tan^4 x – \tan^2 x = 1\) है, तो \(\sin^4 x + \sin^2 x\) का मान ज्ञात कीजिए।

• \(\dfrac{3}{4}\)
• \(\dfrac{1}{2}\)
• \(1\)
• \(\dfrac{3}{2}\)

Explanation:

• माना \(\tan^2 x = t\): \(t^2-t = 1 \Rightarrow t^2 = t+1\)
  \(\sin^2 x = \frac{t}{1+t}\)
  \(\sin^4 x+\sin^2 x = \frac{t^2}{(1+t)^2}+\frac{t}{1+t} = \frac{t^2+t+t^2}{(1+t)^2} = \frac{2t^2+t}{(1+t)^2}\)
  \(t^2 = t+1\) रखने पर अंश \(= 3t+2\), हर \(= t^2+2t+1 = 3t+2\)
  मान \(= 1\)

यदि \(2\sin^2 x = 2 – 3\sin x\) है, तो \(x\) का मान क्या होगा?

• \(\dfrac{\pi}{6}\)
• \(\dfrac{\pi}{4}\)
• \(\dfrac{\pi}{2}\)
• \(\dfrac{\pi}{3}\)

Explanation:

• \(2\sin^2 x = 2-3\sin x \Rightarrow 2\sin^2 x+3\sin x-2 = 0\)
  \((2\sin x-1)(\sin x+2) = 0 \Rightarrow \sin x = \frac{1}{2}\) (\(\sin x = -2\) असंभव)
  \(x = \frac{\pi}{6}\)

यदि \(\dfrac{\sin x + \cos x}{\sin x – \cos x} = \dfrac{6}{5}\) तो \(\dfrac{\tan^2 x + 1}{\tan^2 x – 1}\) का मान ज्ञात करें।

• \(\dfrac{61}{35}\)
• \(\dfrac{61}{60}\)
• \(\dfrac{35}{61}\)
• \(\dfrac{60}{61}\)

Explanation:

• \(\frac{\sin x+\cos x}{\sin x-\cos x} = \frac{6}{5}\); \(\cos x\) से भाग देने पर \(\frac{\tan x+1}{\tan x-1} = \frac{6}{5}\)
  \(5\tan x+5 = 6\tan x-6 \Rightarrow \tan x = 11\)
  \(\frac{\tan^2 x+1}{\tan^2 x-1} = \frac{122}{120} = \frac{61}{60}\)

यदि \(0 < \theta < 90^\circ,\ 3b\csc\theta = a\sec\theta\) और \(3a\sec\theta – b\csc\theta = 8\) है, तो \(9b^2 + a^2\) का मान ज्ञात करें।

• \(6\)
• \(8\)
• \(9\)
• \(7\)

Explanation:

• माना \(a\sec\theta = X\) और \(b\csc\theta = Y\)
  \(3Y = X\) और \(3X-Y = 8 \Rightarrow 9Y-Y = 8 \Rightarrow Y = 1,\ X = 3\)
  \(a = 3\cos\theta\), \(b = \sin\theta\)
  \(9b^2+a^2 = 9\sin^2\theta+9\cos^2\theta = 9\)

यदि \(x\sin^3\theta + y\cos^3\theta = \sin\theta\cos\theta\) और \(x\sin\theta = y\cos\theta\) है, तो \(x^2 + y^2\) का मान ज्ञात करें।

• \(0\)
• \(4\)
• \(1\)
• \(2\)

Explanation:

• \(x\sin^3\theta+y\cos^3\theta = (x\sin\theta)\sin^2\theta+(y\cos\theta)\cos^2\theta\)
  \(x\sin\theta = y\cos\theta\) रखने पर \(= (x\sin\theta)(\sin^2\theta+\cos^2\theta) = x\sin\theta\)
  \(x\sin\theta = \sin\theta\cos\theta \Rightarrow x = \cos\theta\)
  \(y\cos\theta = x\sin\theta = \cos\theta\sin\theta \Rightarrow y = \sin\theta\)
  \(x^2+y^2 = \cos^2\theta+\sin^2\theta = 1\)

यदि \(\sec\theta + \tan\theta = 2 + \sqrt{5}\) और \(\theta\) एक न्यून कोण है, तो \(\sin\theta\) का मान है:

• \(\dfrac{2\sqrt{5}}{5}\)
• \(\dfrac{3}{5}\)
• \(\dfrac{\sqrt{5}}{5}\)
• \(\dfrac{1}{5}\)

Explanation:

• \(\sec\theta+\tan\theta = 2+\sqrt{5} \Rightarrow \sec\theta-\tan\theta = \frac{1}{2+\sqrt{5}} = \sqrt{5}-2\)
  जोड़ने पर \(2\sec\theta = 2\sqrt{5} \Rightarrow \sec\theta = \sqrt{5}\); घटाने पर \(2\tan\theta = 4 \Rightarrow \tan\theta = 2\)
  \(\sin\theta = \frac{\tan\theta}{\sec\theta} = \frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{5}\)

यदि \(2\cos^2\theta = 3(1 – \sin\theta),\ 0^\circ < \theta < 90^\circ\) है, तो \((\tan 2\theta + \csc 3\theta – \sec 2\theta)\) का मान ज्ञात करें।

• \(\sqrt{3} + 1\)
• \(\sqrt{3} – 1\)
• \(\dfrac{1+\sqrt{3}}{\sqrt{3}}\)
• \(\dfrac{1-\sqrt{3}}{\sqrt{3}}\)

Explanation:

• \(2\cos^2\theta = 3(1-\sin\theta) \Rightarrow 2(1-\sin\theta)(1+\sin\theta) = 3(1-\sin\theta)\)
  \(2(1+\sin\theta) = 3 \Rightarrow \sin\theta = \frac{1}{2} \Rightarrow \theta = 30^\circ\)
  \(\tan 60^\circ+\csc 90^\circ-\sec 60^\circ = \sqrt{3}+1-2 = \sqrt{3}-1\)

\(\theta\) का मान ज्ञात करें, यदि \(\sec^2\theta + (1 – \sqrt{3})\tan\theta – (1 + \sqrt{3}) = 0\) जहां, \(\theta\) न्यून कोण है।

• \(60^\circ\)
• \(30^\circ\)
• \(45^\circ\)
• \(15^\circ\)

Explanation:

• \(\sec^2\theta = 1+\tan^2\theta\); माना \(\tan\theta = t\)
  \(1+t^2+(1-\sqrt{3})t-1-\sqrt{3} = 0 \Rightarrow t^2+(1-\sqrt{3})t-\sqrt{3} = 0\)
  \((t+1)(t-\sqrt{3}) = 0 \Rightarrow t = \sqrt{3}\) (न्यून कोण के लिए \(t > 0\))
  \(\theta = 60^\circ\)

यदि \(\csc^2\theta(\cos\theta – 1)(1 + \cos\theta) = k\) हो, तो \(k\) का मान ज्ञात करें।

• \(1\)
• \(-1\)
• \(1/2\)
• \(0\)

Explanation:

• \(\csc^2\theta(\cos\theta-1)(1+\cos\theta) = \csc^2\theta(\cos^2\theta-1) = \csc^2\theta\times(-\sin^2\theta)\)
  \(= -1\)

यदि \(\dfrac{\sin^2\theta}{\tan^2\theta – \sin^2\theta} = 5\) है, तो \(\dfrac{24\cos^2\theta – 15\sec^2\theta}{6\csc^2\theta – 7\cot^2\theta}\) का मान ज्ञात करें।

• \(6\)
• \(2\)
• \(4\)
• \(1\)

Explanation:

• \(\frac{\sin^2\theta}{\tan^2\theta-\sin^2\theta} = \frac{\sin^2\theta}{\sin^2\theta(\sec^2\theta-1)} = \frac{1}{\tan^2\theta} = \cot^2\theta = 5\)
  \(\csc^2\theta = 1+5 = 6\), \(\sec^2\theta = 1+\tan^2\theta = \frac{6}{5} \Rightarrow \cos^2\theta = \frac{5}{6}\)
  अंश \(= 24\times \frac{5}{6}-15\times \frac{6}{5} = 20-18 = 2\)
  हर \(= 6\times 6-7\times 5 = 36-35 = 1\)
  मान \(= 2\)

निम्न में का मान ज्ञात करें: \(3\csc\theta + 4\sin\theta – 4\sqrt{3} = 0\), जहां \(\theta\) न्यून कोण है।

• \(45^\circ\)
• \(30^\circ\)
• \(60^\circ\)
• \(15^\circ\)

Explanation:

• \(3\csc\theta+4\sin\theta = 4\sqrt{3}\); \(\sin\theta\) से गुणा करने पर \(3+4\sin^2\theta = 4\sqrt{3}\sin\theta\)
  \(4\sin^2\theta-4\sqrt{3}\sin\theta+3 = 0 \Rightarrow (2\sin\theta-\sqrt{3})^2 = 0\)
  \(\sin\theta = \frac{\sqrt{3}}{2} \Rightarrow \theta = 60^\circ\)

यदि \(\tan A + \sec A = \dfrac{3}{2}\) है और \(A\) न्यून कोण है, तो \(\dfrac{10\cot A + 13\cos A}{12\tan A + 5\csc A}\) का मान ज्ञात करें।

• \(5\)
• \(1\)
• \(4\)
• \(2\)

Explanation:

• \(\tan A+\sec A = \frac{3}{2} \Rightarrow \sec A-\tan A = \frac{2}{3}\)
  \(\sec A = \frac{1}{2}\left(\frac{3}{2}+\frac{2}{3}\right) = \frac{13}{12}\), \(\tan A = \frac{1}{2}\left(\frac{3}{2}-\frac{2}{3}\right) = \frac{5}{12}\)
  \(\sin A = \frac{5}{13},\ \cos A = \frac{12}{13},\ \cot A = \frac{12}{5},\ \csc A = \frac{13}{5}\)
  अंश \(= 10\times \frac{12}{5}+13\times \frac{12}{13} = 24+12 = 36\)
  हर \(= 12\times \frac{5}{12}+5\times \frac{13}{5} = 5+13 = 18\)
  मान \(= \frac{36}{18} = 2\)

यदि \(\tan\theta + \cot\theta = 4\) है, तो \(3(\tan^2\theta + \cot^2\theta)\) का \((2\csc^2\theta\sec^2\theta – 4)\) से अनुपात ज्ञात करें।

• \(4:3\)
• \(3:4\)
• \(5:4\)
• \(3:2\)

Explanation:

• \(\tan\theta+\cot\theta = 4 \Rightarrow \tan^2\theta+\cot^2\theta = 16-2 = 14\); पहला पद \(= 3\times 14 = 42\)
  \(\tan\theta+\cot\theta = \frac{1}{\sin\theta\cos\theta} = 4 \Rightarrow \csc^2\theta\sec^2\theta = \frac{1}{\sin^2\theta\cos^2\theta} = 16\)
  दूसरा पद \(= 2\times 16-4 = 28\)
  अनुपात \(= 42 : 28 = 3 : 2\)

यदि \(\sin\theta(2\sin\theta + 3) = 2,\ 0^\circ < \theta < 90^\circ\), तो \((\sec^2\theta + \cot^2\theta – \cos^2\theta)\) का मान ज्ञात करें।

• \(\dfrac{13}{3}\)
• \(13\)
• \(\dfrac{7}{2}\)
• \(\dfrac{43}{12}\)

Explanation:

• \(\sin\theta(2\sin\theta+3) = 2 \Rightarrow 2\sin^2\theta+3\sin\theta-2 = 0 \Rightarrow (2\sin\theta-1)(\sin\theta+2) = 0\)
  \(\sin\theta = \frac{1}{2} \Rightarrow \theta = 30^\circ\)
  \(\sec^2\theta = \frac{4}{3},\ \cot^2\theta = 3,\ \cos^2\theta = \frac{3}{4}\)
  मान \(= \frac{4}{3}+3-\frac{3}{4} = \frac{16+36-9}{12} = \frac{43}{12}\)

निम्न समीकरण का मान ज्ञात करें। \(\theta:\ 2\sqrt{3}\sin^2\theta + \cos\theta – \sqrt{3} = 0\) जहां \(\theta\) न्यून कोण है।

• \(30^\circ\)
• \(45^\circ\)
• \(60^\circ\)
• \(15^\circ\)

Explanation:

• \(\sin^2\theta = 1-\cos^2\theta\) रखने पर \(2\sqrt{3}(1-\cos^2\theta)+\cos\theta-\sqrt{3} = 0\)
  \(2\sqrt{3}\cos^2\theta-\cos\theta-\sqrt{3} = 0\)
  \(\cos\theta = \frac{1\pm\sqrt{1+24}}{4\sqrt{3}} = \frac{1\pm 5}{4\sqrt{3}}\)
  न्यून कोण के लिए \(\cos\theta = \frac{6}{4\sqrt{3}} = \frac{\sqrt{3}}{2} \Rightarrow \theta = 30^\circ\)

यदि \(\sec\left(90^\circ – \dfrac{3\theta}{2}\right) = \sqrt{2},\ 0^\circ < \theta < 90^\circ\) है, तो \(2\sin\theta + 4\cos 2\theta\) का मान ज्ञात करें।

• \(4\)
• \(2\)
• \(3\)
• \(6\)

Explanation:

• \(\sec(90^\circ-\frac{3\theta}{2}) = \csc\frac{3\theta}{2} = \sqrt{2} \Rightarrow \sin\frac{3\theta}{2} = \frac{1}{\sqrt{2}}\)
  \(\frac{3\theta}{2} = 45^\circ \Rightarrow \theta = 30^\circ\)
  \(2\sin 30^\circ+4\cos 60^\circ = 2\times \frac{1}{2}+4\times \frac{1}{2} = 1+2 = 3\)

यदि \(\sin x – 3\cos x = \sqrt{3}\cos x\) है, तो \(\cot x\) का मान \(\_\_\_\_\_\_\) है।

• \(3 – \sqrt{3}\)
• \(3 + \sqrt{3}\)
• \(\sqrt{3}\)
• \(\dfrac{3-\sqrt{3}}{6}\)

Explanation:

• \(\sin x-3\cos x = \sqrt{3}\cos x \Rightarrow \sin x = (3+\sqrt{3})\cos x\)
  \(\cot x = \frac{\cos x}{\sin x} = \frac{1}{3+\sqrt{3}} = \frac{3-\sqrt{3}}{(3+\sqrt{3})(3-\sqrt{3})} = \frac{3-\sqrt{3}}{6}\)

यदि \(1 + \tan\theta = \sqrt{3}\), तो \(\sqrt{3}\cot\theta – 1 = \_\_\_\_\_\_\) होगा।

• \(\dfrac{\sqrt{3}+1}{2}\)
• \(\dfrac{\sqrt{3}-1}{2}\)
• \(\dfrac{2\sqrt{3}-1}{2}\)
• \(\dfrac{2\sqrt{3}+1}{2}\)

Explanation:

• \(1+\tan\theta = \sqrt{3} \Rightarrow \tan\theta = \sqrt{3}-1\)
  \(\cot\theta = \frac{1}{\sqrt{3}-1} = \frac{\sqrt{3}+1}{2}\)
  \(\sqrt{3}\cot\theta-1 = \frac{\sqrt{3}(\sqrt{3}+1)}{2}-1 = \frac{3+\sqrt{3}}{2}-1 = \frac{1+\sqrt{3}}{2}\)

यदि \(\csc^2\theta + \cot^2\theta = \dfrac{5}{3}\) है, तो \(\cot 2\theta\) का मान ज्ञात कीजिए।

• \(\dfrac{1}{\sqrt{3}}\)
• \(-\dfrac{1}{\sqrt{3}}\)
• \(\dfrac{2}{\sqrt{3}}\)
• \(-\dfrac{2}{\sqrt{3}}\)

Explanation:

• \(\csc^2\theta = 1+\cot^2\theta\) रखने पर \(1+2\cot^2\theta = \frac{5}{3} \Rightarrow \cot^2\theta = \frac{1}{3}\)
  \(\cot\theta = \frac{1}{\sqrt{3}} \Rightarrow \theta = 60^\circ\)
  \(\cot 2\theta = \cot 120^\circ = -\cot 60^\circ = -\frac{1}{\sqrt{3}}\)

\((\csc\theta – \cot\theta)^2 = \_\_\_\_\_,\ 0^\circ < \theta < 90^\circ\)

• \(\dfrac{1-\sin\theta}{1+\cos\theta}\)
• \(\dfrac{1-\cos\theta}{1+\cos\theta}\)
• \(\dfrac{1+\sin\theta}{1-\cos\theta}\)
• \(\dfrac{1+\cos\theta}{1-\sin\theta}\)

Explanation:

• \((\csc\theta-\cot\theta)^2 = \left(\frac{1-\cos\theta}{\sin\theta}\right)^2 = \frac{(1-\cos\theta)^2}{1-\cos^2\theta}\)
  \(= \frac{(1-\cos\theta)^2}{(1-\cos\theta)(1+\cos\theta)} = \frac{1-\cos\theta}{1+\cos\theta}\)

यदि \(\sin\theta + \cos\theta = \sqrt{2}\) है, तो \(\csc\theta + \cot\theta\) का मान ज्ञात कीजिए।

• \(\sqrt{2} – 1\)
• \(\sqrt{2} + 5\)
• \(\sqrt{3} – 1\)
• \(\sqrt{2} + 1\)

Explanation:

• \((\sin\theta+\cos\theta)^2 = 2 \Rightarrow 1+2\sin\theta\cos\theta = 2 \Rightarrow \sin\theta\cos\theta = \frac{1}{2}\)
  \((\sin\theta-\cos\theta)^2 = 1-1 = 0 \Rightarrow \sin\theta = \cos\theta = \frac{1}{\sqrt{2}}\) (\(\theta = 45^\circ\))
  \(\csc\theta+\cot\theta = \sqrt{2}+1\)

यदि \((2\cos^3 A – \cos A)/(\sin A – 2\sin^3 A) = 1\) तो \(A\) का मान क्या है?

• \(90^\circ\)
• \(30^\circ\)
• \(45^\circ\)
• \(60^\circ\)

Explanation:

• अंश \(= \cos A(2\cos^2 A-1) = \cos A\cos 2A\); हर \(= \sin A(1-2\sin^2 A) = \sin A\cos 2A\)
  \(\frac{\cos A}{\sin A} = \cot A = 1 \Rightarrow A = 45^\circ\)

यदि \(\sec\beta + \tan\beta = 2\), तो \(\cot\beta\) का मान क्या है?

• \(\dfrac{5}{3}\)
• \(\dfrac{3}{5}\)
• \(\dfrac{4}{3}\)
• \(\dfrac{3}{4}\)

Explanation:

• \(\sec\beta+\tan\beta = 2 \Rightarrow \sec\beta-\tan\beta = \frac{1}{2}\)
  \(\sec\beta = \frac{1}{2}\left(2+\frac{1}{2}\right) = \frac{5}{4}\), \(\tan\beta = \frac{1}{2}\left(2-\frac{1}{2}\right) = \frac{3}{4}\)
  \(\cot\beta = \frac{4}{3}\)

यदि \(\sec x + \cos x = \dfrac{5}{2}\), जहां \(x,\ 0^\circ\) और \(90^\circ\) के मध्य स्थित है, तो \(\sin^2 x\) का मान क्या है?

• \(\dfrac{3}{4}\)
• \(\dfrac{1}{2}\)
• \(1\)
• \(\dfrac{1}{4}\)

Explanation:

• माना \(\sec x = s\): \(s+\frac{1}{s} = \frac{5}{2} \Rightarrow 2s^2-5s+2 = 0 \Rightarrow (2s-1)(s-2) = 0\)
  \(\sec x \ge 1\) इसलिए \(\sec x = 2 \Rightarrow \cos x = \frac{1}{2}\)
  \(\sin^2 x = 1-\frac{1}{4} = \frac{3}{4}\)

\(\dfrac{\sin\theta + \cos\theta}{\sin\theta – \cos\theta} + \dfrac{\sin\theta – \cos\theta}{\sin\theta + \cos\theta} = ?\)

• \(\dfrac{1}{(\sin^2\theta – \cos^2\theta)}\)
• \(2(\sin^2\theta – \cos^2\theta)\)
• \(\dfrac{2}{(\sin^2\theta – \cos^2\theta)}\)
• \((\sin^2\theta – \cos^2\theta)\)

Explanation:

• सामान्य हर \(= (\sin\theta-\cos\theta)(\sin\theta+\cos\theta) = \sin^2\theta-\cos^2\theta\)
  अंश \(= (\sin\theta+\cos\theta)^2+(\sin\theta-\cos\theta)^2 = 2(\sin^2\theta+\cos^2\theta) = 2\)
  मान \(= \frac{2}{\sin^2\theta-\cos^2\theta}\)

यदि \(\cos\theta + \sin\theta = \sqrt{2}\) है, तो \(\sec\theta\csc\theta\) का मान क्या है?

• \(\dfrac{1}{2}\)
• \(1\)
• \(2\)
• \(0\)

Explanation:

• \((\cos\theta+\sin\theta)^2 = 2 \Rightarrow 1+2\sin\theta\cos\theta = 2 \Rightarrow \sin\theta\cos\theta = \frac{1}{2}\)
  \(\sec\theta\csc\theta = \frac{1}{\sin\theta\cos\theta} = 2\)

यदि \(\sin\alpha + \cos\alpha = \dfrac{2}{\sqrt{3}}\) है, तो \((\tan\alpha + \cot\alpha)\) का मान ज्ञात करें?

• \(0\)
• \(2\)
• \(6\)
• \(1\)

Explanation:

• \((\sin\alpha+\cos\alpha)^2 = \frac{4}{3} \Rightarrow 1+2\sin\alpha\cos\alpha = \frac{4}{3} \Rightarrow \sin\alpha\cos\alpha = \frac{1}{6}\)
  \(\tan\alpha+\cot\alpha = \frac{1}{\sin\alpha\cos\alpha} = 6\)

\(\dfrac{2\sec A}{\sec^2 A – 1} = ?\)

• \(2\csc A\cot A\)
• \(2\cos A\cot A\)
• \(2\csc A\tan A\)
• \(2\cos A\sin A\)

Explanation:

• \(\sec^2 A-1 = \tan^2 A\)
  \(\frac{2\sec A}{\tan^2 A} = \frac{2}{\cos A}\times \frac{\cos^2 A}{\sin^2 A} = \frac{2\cos A}{\sin^2 A} = 2\times \frac{1}{\sin A}\times \frac{\cos A}{\sin A} = 2\csc A\cot A\)

यदि \(\cos^4\theta – \sin^4\theta = \dfrac{4}{5}\) है, तो \(\sin 4\theta\) ज्ञात कीजिये।

• \(\dfrac{24}{25}\)
• \(\dfrac{21}{25}\)
• \(\dfrac{16}{25}\)
• \(\dfrac{18}{25}\)

Explanation:

• \(\cos^4\theta-\sin^4\theta = (\cos^2\theta-\sin^2\theta)(\cos^2\theta+\sin^2\theta) = \cos 2\theta = \frac{4}{5}\)
  \(\sin 2\theta = \sqrt{1-\frac{16}{25}} = \frac{3}{5}\)
  \(\sin 4\theta = 2\sin 2\theta\cos 2\theta = 2\times \frac{3}{5}\times \frac{4}{5} = \frac{24}{25}\)

यदि \(\tan A = a\tan B\) और \(\sin A = b\sin B\), तो \(\cos A\) का मान है:

• \(\dfrac{b^2-1}{a^2-1}\)
• \(\dfrac{a^2-1}{b^2-1}\)
• \(\dfrac{b^2+1}{a^2+1}\)
• \(\dfrac{a^2+1}{b^2+1}\)

Explanation:

• \(\frac{\tan A}{\sin A} = \frac{a\tan B}{b\sin B} \Rightarrow \frac{1}{\cos A} = \frac{a}{b\cos B} \Rightarrow \cos B = \frac{a}{b}\cos A\)
  \(\sin^2 A = b^2\sin^2 B = b^2\left(1-\frac{a^2}{b^2}\cos^2 A\right) = b^2-a^2\cos^2 A\)
  \(1-\cos^2 A = b^2-a^2\cos^2 A \Rightarrow (a^2-1)\cos^2 A = b^2-1\)
  \(\cos^2 A = \frac{b^2-1}{a^2-1}\)
  नोट: यह \(\cos^2 A\) का मान है; विकल्प/कुंजी में इसी को \(\cos A\) के स्थान पर लिखा गया है।

\((\cot\theta + \tan\theta)(\csc\theta – \sin\theta)(\cos\theta – \sec\theta) = ?\)

• \(0\)
• \(2\)
• \(1\)
• \(-1\)

Explanation:

• \(\cot\theta+\tan\theta = \frac{1}{\sin\theta\cos\theta}\); \(\csc\theta-\sin\theta = \frac{\cos^2\theta}{\sin\theta}\); \(\cos\theta-\sec\theta = -\frac{\sin^2\theta}{\cos\theta}\)
  गुणनफल \(= \frac{1}{\sin\theta\cos\theta}\times \frac{\cos^2\theta}{\sin\theta}\times \left(-\frac{\sin^2\theta}{\cos\theta}\right) = -1\)

\(\dfrac{3\cos^3\theta – 2\cos\theta}{\sin\theta – 3\sin^3\theta}\) बराबर है:

• \(\tan\theta\)
• \(\cot^2\theta\)
• \(\cot\theta\)
• \(\tan^2\theta\)

Explanation:

• अंश \(= \cos\theta(3\cos^2\theta-2) = \cos\theta(3-3\sin^2\theta-2) = \cos\theta(1-3\sin^2\theta)\)
  हर \(= \sin\theta(1-3\sin^2\theta)\)
  मान \(= \frac{\cos\theta}{\sin\theta} = \cot\theta\)

यदि \(12\sin^2 A + 2\cos^2 A = 7\), जहाँ \(0^\circ < A < 180^\circ\), तो \(\cot A\) का मान ज्ञात कीजिये।

• \(\pm 1\)
• \(\pm\dfrac{1}{2}\)
• \(\pm\dfrac{1}{\sqrt{2}}\)
• \(\pm\dfrac{1}{\sqrt{3}}\)

Explanation:

• \(12\sin^2 A+2\cos^2 A = 7 \Rightarrow 2(\sin^2 A+\cos^2 A)+10\sin^2 A = 7\)
  \(2+10\sin^2 A = 7 \Rightarrow \sin^2 A = \frac{1}{2} \Rightarrow \cos^2 A = \frac{1}{2}\)
  \(\cot^2 A = 1 \Rightarrow \cot A = \pm 1\) (\(0^\circ < A < 180^\circ\) में दोनों संभव)

यदि \(4\sin^2\theta – 4\sqrt{3}\sin\theta + 3 = 0\), तो \(\theta\) का मान ज्ञात कीजिये।

• \(30^\circ\)
• \(75^\circ\)
• \(60^\circ\)
• \(45^\circ\)

Explanation:

• \(4\sin^2\theta-4\sqrt{3}\sin\theta+3 = 0 \Rightarrow (2\sin\theta-\sqrt{3})^2 = 0\)
  \(\sin\theta = \frac{\sqrt{3}}{2} \Rightarrow \theta = 60^\circ\)

यदि \(\theta\) एक न्यून कोण है और \(\tan\theta + \cot\theta = 2\) है, तो \(\tan^{12}\theta + \cot^{12}\theta + 2\tan^5\theta\cot^7\theta\) ज्ञात कीजिये।

• \(4\)
• \(3\)
• \(2\)
• \(1\)

Explanation:

• \(\tan\theta+\cot\theta = 2 \Rightarrow (\tan\theta-1)^2 = 0 \Rightarrow \tan\theta = \cot\theta = 1\)
  \(\tan^{12}\theta+\cot^{12}\theta+2\tan^5\theta\cot^7\theta = 1+1+2 = 4\)

अगर \((\cos^2\theta – 1)(2\sec^2\theta) + \sec^2\theta + 2\tan^2\theta = 2,\ 0^\circ < \theta < 90^\circ\) है, तो \(\dfrac{(\csc\theta + \cos\theta)}{(\csc\theta – \cos\theta)}\) का मान \(\_\_\_\_\_\) होगा।

• \(4\sqrt{2}\)
• \(1\)
• \(3\)
• \(2\sqrt{2}\)

Explanation:

• \((\cos^2\theta-1)\times 2\sec^2\theta = -\sin^2\theta\times \frac{2}{\cos^2\theta} = -2\tan^2\theta\)
  \(-2\tan^2\theta+\sec^2\theta+2\tan^2\theta = \sec^2\theta = 2 \Rightarrow \cos^2\theta = \frac{1}{2} \Rightarrow \theta = 45^\circ\)
  \(\frac{\csc\theta+\cos\theta}{\csc\theta-\cos\theta} = \frac{\sqrt{2}+\frac{1}{\sqrt{2}}}{\sqrt{2}-\frac{1}{\sqrt{2}}} = \frac{3/\sqrt{2}}{1/\sqrt{2}} = 3\)

यदि \(\cos^2\theta – \sin^2\theta – 3\cos\theta + 2 = 0,\ 0^\circ < \theta < 90^\circ\), तो \(5\cos\theta – \dfrac{\tan\theta}{2}\) का मान क्या होगा?

• \(\dfrac{5+\sqrt{3}}{2}\)
• \(\dfrac{3-\sqrt{3}}{2}\)
• \(\dfrac{5-\sqrt{3}}{2}\)
• \(\dfrac{3+\sqrt{3}}{2}\)

Explanation:

• \(\cos^2\theta-\sin^2\theta-3\cos\theta+2 = 0 \Rightarrow 2\cos^2\theta-3\cos\theta+1 = 0\)
  \((2\cos\theta-1)(\cos\theta-1) = 0 \Rightarrow \cos\theta = \frac{1}{2}\) (\(\cos\theta = 1\) मान्य नहीं) \(\Rightarrow \theta = 60^\circ\)
  \(5\cos\theta-\frac{\tan\theta}{2} = \frac{5}{2}-\frac{\sqrt{3}}{2} = \frac{5-\sqrt{3}}{2}\)

यदि \(3\tan\theta = 2\sqrt{3}\sin\theta,\ 0^\circ < \theta < 90^\circ\) है, तो \(\dfrac{3}{4}\left(\dfrac{\csc^2 2\theta + \cot^2 2\theta}{\sin^2\theta + \tan^2 2\theta}\right)\) का मान ज्ञात करें।

• \(\dfrac{5}{13}\)
• \(\dfrac{3}{13}\)
• \(\dfrac{7}{13}\)
• \(\dfrac{1}{13}\)

Explanation:

• \(3\tan\theta = 2\sqrt{3}\sin\theta \Rightarrow \frac{3}{\cos\theta} = 2\sqrt{3} \Rightarrow \cos\theta = \frac{\sqrt{3}}{2} \Rightarrow \theta = 30^\circ\), \(2\theta = 60^\circ\)
  \(\csc^2 60^\circ+\cot^2 60^\circ = \frac{4}{3}+\frac{1}{3} = \frac{5}{3}\)
  \(\sin^2 30^\circ+\tan^2 60^\circ = \frac{1}{4}+3 = \frac{13}{4}\)
  मान \(= \frac{3}{4}\times \frac{5/3}{13/4} = \frac{3}{4}\times \frac{20}{39} = \frac{5}{13}\)

यदि \(\dfrac{1}{1+\tan\theta} + \dfrac{1}{1-\tan\theta} = 4,\ 0^\circ < \theta < 90^\circ\), तो \(\csc\theta + \sec\theta + \sin\theta\) का मान क्या है?

• \(\dfrac{3+4\sqrt{2}}{\sqrt{6}}\)
• \(\dfrac{8\sqrt{3}}{3}\)
• \(\dfrac{3\sqrt{3}}{2}\)
• \(\dfrac{4+3\sqrt{2}}{\sqrt{6}}\)

Explanation:

• \(\frac{1}{1+\tan\theta}+\frac{1}{1-\tan\theta} = \frac{2}{1-\tan^2\theta} = 4 \Rightarrow 1-\tan^2\theta = \frac{1}{2} \Rightarrow \tan\theta = \frac{1}{\sqrt{2}}\)
  \(\sin\theta = \frac{1}{\sqrt{3}},\ \cos\theta = \frac{\sqrt{2}}{\sqrt{3}}\) \(\Rightarrow \csc\theta = \sqrt{3},\ \sec\theta = \frac{\sqrt{3}}{\sqrt{2}}\)
  मान \(= \sqrt{3}+\frac{\sqrt{3}}{\sqrt{2}}+\frac{1}{\sqrt{3}} = \frac{4}{\sqrt{3}}+\frac{\sqrt{3}}{\sqrt{2}} = \frac{4\sqrt{2}+3}{\sqrt{6}}\)

यदि \(14\sin^2\theta + 3\sin\theta – 5 = 0,\ 0^\circ < \theta < 90^\circ\), तो \(\dfrac{\csc\theta + \tan\theta}{\sec\theta + \cot\theta}\) का मान क्या होगा?

• \(\dfrac{3\sqrt{3}-1}{3}\)
• \(\dfrac{2\sqrt{3}+1}{5}\)
• \(\dfrac{2(1+\sqrt{3})}{5}\)
• \(\dfrac{1+2\sqrt{3}}{5}\)

Explanation:

• \(14\sin^2\theta+3\sin\theta-5 = 0 \Rightarrow \sin\theta = \frac{-3+\sqrt{9+280}}{28} = \frac{-3+17}{28} = \frac{1}{2}\)
  \(\theta = 30^\circ\)
  \(\frac{\csc\theta+\tan\theta}{\sec\theta+\cot\theta} = \frac{2+\frac{1}{\sqrt{3}}}{\frac{2}{\sqrt{3}}+\sqrt{3}} = \frac{\frac{2\sqrt{3}+1}{\sqrt{3}}}{\frac{5}{\sqrt{3}}} = \frac{2\sqrt{3}+1}{5}\)

यदि \(\sin\theta = \sqrt{3}\cos\theta,\ 0^\circ < \theta < 90^\circ\) है, तो \(2\sin^2\theta + \sec^2\theta + \sin\theta\sec\theta + \csc\theta\) का मान है?

• \(\dfrac{33+10\sqrt{3}}{3}\)
• \(\dfrac{19+10\sqrt{3}}{3}\)
• \(\dfrac{19+10\sqrt{3}}{6}\)
• \(\dfrac{33+10\sqrt{3}}{6}\)

Explanation:

• \(\sin\theta = \sqrt{3}\cos\theta \Rightarrow \tan\theta = \sqrt{3} \Rightarrow \theta = 60^\circ\)
  \(2\sin^2\theta = \frac{3}{2}\), \(\sec^2\theta = 4\), \(\sin\theta\sec\theta = \sqrt{3}\), \(\csc\theta = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}\)
  कुल \(= \frac{3}{2}+4+\sqrt{3}+\frac{2\sqrt{3}}{3} = \frac{11}{2}+\frac{5\sqrt{3}}{3} = \frac{33+10\sqrt{3}}{6}\)

\(\left(\dfrac{1-\tan\theta}{1-\cot\theta}\right)^2 + 1 = ?\)

• \(\sec^2\theta\)
• \(\cos^2\theta\)
• \(\sin^2\theta\)
• \(\csc^2\theta\)

Explanation:

• \(\frac{1-\tan\theta}{1-\cot\theta} = \frac{1-\tan\theta}{1-\frac{1}{\tan\theta}} = \frac{1-\tan\theta}{\frac{\tan\theta-1}{\tan\theta}} = -\tan\theta\)
  \((-\tan\theta)^2+1 = \tan^2\theta+1 = \sec^2\theta\)

\(\left(\dfrac{\sin A}{1-\cos A} + \dfrac{1-\cos A}{\sin A}\right) \div \left(\dfrac{\cot^2 A}{1+\csc A} + 1\right)\) का मान है:

• \(3/2\)
• \(1/2\)
• \(1\)
• \(2\)

Explanation:

• पहला कोष्ठक: \(\frac{\sin^2 A+(1-\cos A)^2}{\sin A(1-\cos A)} = \frac{2-2\cos A}{\sin A(1-\cos A)} = \frac{2}{\sin A} = 2\csc A\)
  दूसरा कोष्ठक: \(\frac{\cot^2 A}{1+\csc A}+1 = \frac{\csc^2 A-1}{1+\csc A}+1 = (\csc A-1)+1 = \csc A\)
  मान \(= \frac{2\csc A}{\csc A} = 2\)

\(\dfrac{\sec\theta(1-\sin\theta)(\sin\theta+\cos\theta)(\sec\theta+\tan\theta)}{\sin\theta(1+\tan\theta)+\cos\theta(1+\cot\theta)} = ?\)

• \(2\cos\theta\)
• \(\csc\theta\cdot\sec\theta\)
• \(2\sin\theta\)
• \(\sin\theta\cdot\cos\theta\)

Explanation:

• अंश: \(\sec\theta(1-\sin\theta)(\sec\theta+\tan\theta) = \sec\theta(1-\sin\theta)\cdot\frac{1+\sin\theta}{\cos\theta} = \sec\theta\cos\theta = 1\), इसलिए अंश \(= (\sin\theta+\cos\theta)\)
  हर: \(\sin\theta(1+\tan\theta)+\cos\theta(1+\cot\theta) = \frac{\sin\theta(\cos\theta+\sin\theta)}{\cos\theta}+\frac{\cos\theta(\sin\theta+\cos\theta)}{\sin\theta}\)
  \(= (\sin\theta+\cos\theta)\left(\frac{\sin\theta}{\cos\theta}+\frac{\cos\theta}{\sin\theta}\right) = \frac{\sin\theta+\cos\theta}{\sin\theta\cos\theta}\)
  मान \(= \sin\theta\cos\theta\)

\(\tan^2\theta + \cot^2\theta – \sec^2\theta\csc^2\theta\) का मान निम्नलिखित में से किसके बराबर होगा?

• \(-2\)
• \(1\)
• \(0\)
• \(-1\)

Explanation:

• \(\sec^2\theta\csc^2\theta = (1+\tan^2\theta)(1+\cot^2\theta) = 2+\tan^2\theta+\cot^2\theta\)
  व्यंजक \(= \tan^2\theta+\cot^2\theta-(2+\tan^2\theta+\cot^2\theta) = -2\)

\((\sec\theta – \tan\theta)^2(1 + \sin\theta)^2 + \sin^2\theta = ?\)

• \(\cos\theta\)
• \(\cot^2\theta\)
• \(\sec\theta\)
• \(\cos^2\theta\)

Explanation:

• \((\sec\theta-\tan\theta)(1+\sin\theta) = \frac{(1-\sin\theta)(1+\sin\theta)}{\cos\theta} = \frac{\cos^2\theta}{\cos\theta} = \cos\theta\)
  अतः \((\sec\theta-\tan\theta)^2(1+\sin\theta)^2 = \cos^2\theta\)
  जैसा छपा है: \(\cos^2\theta+\sin^2\theta = 1\)
  नोट: जैसा लिखा है उसका मान 1 आता है, जो विकल्पों में नहीं है। उत्तर कुंजी (\(\cot^2\theta\)) तब मिलती है जब व्यंजक \(\frac{(\sec\theta-\tan\theta)^2(1+\sin\theta)^2}{\sin^2\theta} = \frac{\cos^2\theta}{\sin^2\theta} = \cot^2\theta\) हो।

यदि \(3(\cot^2\theta – \cos^2\theta) = \cos^2\theta,\ 0^\circ < \theta < 90^\circ\) है, तो \((\tan^2\theta + \csc^2\theta + \sin^2\theta)\) का मान है:

• \(\dfrac{13}{3}\)
• \(\dfrac{61}{12}\)
• \(\dfrac{25}{12}\)
• \(\dfrac{15}{4}\)

Explanation:

• \(3(\cot^2\theta-\cos^2\theta) = \cos^2\theta \Rightarrow \cot^2\theta = \frac{4}{3}\cos^2\theta\)
  \(\frac{\cos^2\theta}{\sin^2\theta} = \frac{4}{3}\cos^2\theta \Rightarrow \sin^2\theta = \frac{3}{4} \Rightarrow \theta = 60^\circ\)
  \(\tan^2\theta = 3\), \(\csc^2\theta = \frac{4}{3}\), \(\sin^2\theta = \frac{3}{4}\)
  मान \(= 3+\frac{4}{3}+\frac{3}{4} = \frac{36+16+9}{12} = \frac{61}{12}\)

यदि \(\dfrac{\sin^2\theta – 3\sin\theta + 2}{\cos^2\theta} = 1\) है, जिसमें \(0^\circ < \theta < 90^\circ\) है, तो \((\cos 2\theta + \sin 3\theta + \csc 2\theta)\) का मान क्या है?

• \(\dfrac{2+\sqrt{3}}{3}\)
• \(\dfrac{3+4\sqrt{3}}{6}\)
• \(\dfrac{9+4\sqrt{3}}{6}\)
• \(\dfrac{3+2\sqrt{3}}{3}\)

Explanation:

• \(\sin^2\theta-3\sin\theta+2 = \cos^2\theta = 1-\sin^2\theta \Rightarrow 2\sin^2\theta-3\sin\theta+1 = 0\)
  \((2\sin\theta-1)(\sin\theta-1) = 0 \Rightarrow \sin\theta = \frac{1}{2} \Rightarrow \theta = 30^\circ\)
  \(\cos 60^\circ+\sin 90^\circ+\csc 60^\circ = \frac{1}{2}+1+\frac{2}{\sqrt{3}} = \frac{3}{2}+\frac{2\sqrt{3}}{3} = \frac{9+4\sqrt{3}}{6}\)

\(\dfrac{\sec^2\theta}{\csc^2\theta} + \dfrac{\csc^2\theta}{\sec^2\theta} – (\sec^2\theta + \csc^2\theta) = ?\)

• \(0\)
• \(-2\)
• \(2\)
• \(1\)

Explanation:

• \(\frac{\sec^2\theta}{\csc^2\theta} = \frac{\sin^2\theta}{\cos^2\theta} = \tan^2\theta\) और \(\frac{\csc^2\theta}{\sec^2\theta} = \cot^2\theta\)
  \(\sec^2\theta+\csc^2\theta = 2+\tan^2\theta+\cot^2\theta\)
  व्यंजक \(= \tan^2\theta+\cot^2\theta-2-\tan^2\theta-\cot^2\theta = -2\)

यदि \(\dfrac{\sin\theta}{1+\cos\theta} + \dfrac{1+\cos\theta}{\sin\theta} = \dfrac{4}{\sqrt{3}},\ 0^\circ < \theta < 90^\circ\), तो \((\tan\theta + \sec\theta)^{-1}\) का मान है:

• \(2 – \sqrt{3}\)
• \(3 – \sqrt{2}\)
• \(2 + \sqrt{3}\)
• \(3 + \sqrt{2}\)

Explanation:

• \(\frac{\sin\theta}{1+\cos\theta}+\frac{1+\cos\theta}{\sin\theta} = \frac{\sin^2\theta+(1+\cos\theta)^2}{\sin\theta(1+\cos\theta)} = \frac{2(1+\cos\theta)}{\sin\theta(1+\cos\theta)} = \frac{2}{\sin\theta}\)
  \(\frac{2}{\sin\theta} = \frac{4}{\sqrt{3}} \Rightarrow \sin\theta = \frac{\sqrt{3}}{2} \Rightarrow \theta = 60^\circ\)
  \((\tan 60^\circ+\sec 60^\circ)^{-1} = \frac{1}{\sqrt{3}+2} = 2-\sqrt{3}\)

\(\cos A(\sec A – \cos A)\cdot(\cot A + \tan A) = ?\)

• \(\sec A\)
• \(\cot A\)
• \(\sin A\)
• \(\tan A\)

Explanation:

• \(\sec A-\cos A = \frac{1-\cos^2 A}{\cos A} = \frac{\sin^2 A}{\cos A}\)
  \(\cos A\cdot\frac{\sin^2 A}{\cos A}\cdot(\cot A+\tan A) = \sin^2 A\cdot\frac{1}{\sin A\cos A} = \frac{\sin A}{\cos A} = \tan A\)

\(\dfrac{\sec^2\theta(2+\tan^2\theta+\cot^2\theta) \div (\sin^2\theta – \tan^2\theta)}{(\csc^2\theta + \sec^2\theta)(1+\cot^2\theta)^2} = ?\)

• \(-1\)
• \(1\)
• \(-2\)
• \(2\)

Explanation:

• \(2+\tan^2\theta+\cot^2\theta = \sec^2\theta\csc^2\theta\); \(\sin^2\theta-\tan^2\theta = -\frac{\sin^4\theta}{\cos^2\theta}\)
  अंश \(= \sec^2\theta\times \sec^2\theta\csc^2\theta\times\left(-\frac{\cos^2\theta}{\sin^4\theta}\right) = -\frac{1}{\cos^2\theta\sin^6\theta}\)
  हर \(= (\csc^2\theta+\sec^2\theta)(\csc^2\theta)^2 = \frac{1}{\sin^2\theta\cos^2\theta}\times \frac{1}{\sin^4\theta} = \frac{1}{\sin^6\theta\cos^2\theta}\)
  मान \(= -1\)

\(\dfrac{2\sin^2 38^\circ\sec^2 52^\circ + \cos 64^\circ\sin 26^\circ + \sin^2 64^\circ}{\tan^2 23^\circ + \cot^2 23^\circ – \sec^2 67^\circ – \csc^2 67^\circ}\) का मान है:

• \(-2\)
• \(\dfrac{3}{2}\)
• \(2\)
• \(-\dfrac{3}{2}\)

Explanation:

• \(\sec 52^\circ = \csc 38^\circ \Rightarrow 2\sin^2 38^\circ\sec^2 52^\circ = 2\sin^2 38^\circ\csc^2 38^\circ = 2\)
  \(\sin 26^\circ = \cos 64^\circ \Rightarrow \cos 64^\circ\sin 26^\circ+\sin^2 64^\circ = \cos^2 64^\circ+\sin^2 64^\circ = 1\)
  अंश \(= 2+1 = 3\)
  \(\sec 67^\circ = \csc 23^\circ,\ \csc 67^\circ = \sec 23^\circ\) \(\Rightarrow\) हर \(= \tan^2 23^\circ+\cot^2 23^\circ-\csc^2 23^\circ-\sec^2 23^\circ = -2\)
  मान \(= \frac{3}{-2} = -\frac{3}{2}\)

\((\tan^2 A + \cot^2 A – 2) – \sec^2 A\csc^2 A = ?\)

• \(-4\)
• \(-1\)
• \(1\)
• \(4\)

Explanation:

• \(\sec^2 A\csc^2 A = 2+\tan^2 A+\cot^2 A\)
  व्यंजक \(= \tan^2 A+\cot^2 A-2-(2+\tan^2 A+\cot^2 A) = -4\)

\(\dfrac{\sin^2\theta}{\cos\theta(1+\cos\theta)} + \dfrac{1+\cos\theta}{\cos\theta} = ?\)

• \(2\cos\theta\)
• \(\sec\theta\)
• \(2\sec\theta\)
• \(\csc\theta\)

Explanation:

• \(\frac{\sin^2\theta}{\cos\theta(1+\cos\theta)} = \frac{(1-\cos\theta)(1+\cos\theta)}{\cos\theta(1+\cos\theta)} = \frac{1-\cos\theta}{\cos\theta}\)
  कुल \(= \frac{1-\cos\theta}{\cos\theta}+\frac{1+\cos\theta}{\cos\theta} = \frac{2}{\cos\theta} = 2\sec\theta\)

यदि \(\dfrac{\cos^2\theta}{\cot^2\theta – \cos^2\theta}\) जहाँ \(0^\circ < \theta < 90^\circ\), तो \(\theta\) का मान ज्ञात कीजिए।

• \(60^\circ\)
• \(45^\circ\)
• \(50^\circ\)
• \(30^\circ\)

Explanation:

• \(\frac{\cos^2\theta}{\cot^2\theta-\cos^2\theta} = \frac{\cos^2\theta}{\cos^2\theta\left(\frac{1}{\sin^2\theta}-1\right)} = \frac{1}{\csc^2\theta-1} = \frac{1}{\cot^2\theta} = \tan^2\theta\)
  उत्तर कुंजी (\(60^\circ\)) के लिए \(\tan^2\theta = 3\) होना चाहिए, क्योंकि \(\tan^2 60^\circ = 3\)
  नोट: प्रश्न में “\(= 3\)” छूटा हुआ प्रतीत होता है; व्यंजक \(= 3\) मानने पर \(\tan\theta = \sqrt{3} \Rightarrow \theta = 60^\circ\)।

यदि \((\sin\theta + \csc\theta)^2 + (\cos\theta + \sec\theta)^2 = k + \tan^2\theta + \cot^2\theta\), तो \(k\) का मान ज्ञात कीजिए।

• \(7\)
• \(5\)
• \(9\)
• \(2\)

Explanation:

• \((\sin\theta+\csc\theta)^2 = \sin^2\theta+2+\csc^2\theta\) और \((\cos\theta+\sec\theta)^2 = \cos^2\theta+2+\sec^2\theta\)
  जोड़ने पर \(= 1+4+\csc^2\theta+\sec^2\theta = 5+(1+\cot^2\theta)+(1+\tan^2\theta)\)
  \(= 7+\tan^2\theta+\cot^2\theta\)
  इसलिए \(k = 7\)

यदि \(3\tan\theta = 2\sqrt{3}\sin\theta,\ 0^\circ < \theta < 90^\circ\) है, तो \(\dfrac{\csc^2 2\theta + \cot^2 2\theta}{\sin^2\theta + \tan^2 2\theta}\) का मान क्या होगा?

• \(4/13\)
• \(20/39\)
• \(4/3\)
• \(20/27\)

Explanation:

• \(3\tan\theta = 2\sqrt{3}\sin\theta \Rightarrow \frac{3}{\cos\theta} = 2\sqrt{3} \Rightarrow \cos\theta = \frac{\sqrt{3}}{2} \Rightarrow \theta = 30^\circ\), \(2\theta = 60^\circ\)
  \(\csc^2 60^\circ+\cot^2 60^\circ = \frac{4}{3}+\frac{1}{3} = \frac{5}{3}\)
  \(\sin^2 30^\circ+\tan^2 60^\circ = \frac{1}{4}+3 = \frac{13}{4}\)
  मान \(= \frac{5}{3}\times \frac{4}{13} = \frac{20}{39}\)

यदि \(1 + 2\tan^2\theta + 2\sin\theta\sec^2\theta = \dfrac{a}{b},\ 0^\circ < \theta < 90^\circ\) तो \(\dfrac{a+b}{a-b} = ?\)

• \(\sin\theta\)
• \(\cos\theta\)
• \(\csc\theta\)
• \(\sec\theta\)

Explanation:

• \(1+2\tan^2\theta+2\sin\theta\sec^2\theta = \frac{\cos^2\theta+2\sin^2\theta+2\sin\theta}{\cos^2\theta} = \frac{1+\sin^2\theta+2\sin\theta}{\cos^2\theta} = \frac{(1+\sin\theta)^2}{(1-\sin\theta)(1+\sin\theta)} = \frac{1+\sin\theta}{1-\sin\theta}\)
  \(\frac{a}{b} = \frac{1+\sin\theta}{1-\sin\theta}\); योगांतरानुपात से \(\frac{a+b}{a-b} = \frac{2}{2\sin\theta} = \csc\theta\)

यदि \(7\sin^2\theta + 4\cos^2\theta = 5\) और \(\theta\) पहले चतुर्थांश में स्थित हैं, तो \(\dfrac{\sqrt{3}\sec\theta + \tan\theta}{\sqrt{2}\cot\theta – \sqrt{3}\cos\theta}\) का मान क्या है?

• \(2(1+\sqrt{2})\)
• \(3\sqrt{2}\)
• \(2(\sqrt{2}-1)\)
• \(4\sqrt{2}\)

Explanation:

• \(7\sin^2\theta+4\cos^2\theta = 5 \Rightarrow 4+3\sin^2\theta = 5 \Rightarrow \sin^2\theta = \frac{1}{3}\)
  \(\sin\theta = \frac{1}{\sqrt{3}},\ \cos\theta = \frac{\sqrt{2}}{\sqrt{3}},\ \tan\theta = \frac{1}{\sqrt{2}},\ \sec\theta = \frac{\sqrt{3}}{\sqrt{2}},\ \cot\theta = \sqrt{2}\)
  अंश \(= \sqrt{3}\times \frac{\sqrt{3}}{\sqrt{2}}+\frac{1}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}\)
  हर \(= \sqrt{2}\times \sqrt{2}-\sqrt{3}\times \frac{\sqrt{2}}{\sqrt{3}} = 2-\sqrt{2}\)
  मान \(= \frac{2\sqrt{2}}{2-\sqrt{2}} = \frac{2\sqrt{2}(2+\sqrt{2})}{2} = 2\sqrt{2}+2 = 2(1+\sqrt{2})\)

यदि \(\dfrac{\sin^2\theta}{\cos^2\theta – 3\cos\theta + 2} = 1\), \(\theta\) चतुर्थांश में है, तो \(\dfrac{\tan^2\frac{\theta}{2} + \sin^2\frac{\theta}{2}}{\tan\theta + \sin\theta}\) का मान है?

• \(\dfrac{2\sqrt{3}}{27}\)
• \(\dfrac{5\sqrt{3}}{27}\)
• \(\dfrac{2\sqrt{3}}{9}\)
• \(\dfrac{7\sqrt{3}}{54}\)

Explanation:

• \(\sin^2\theta = 1-\cos^2\theta\): \(1-\cos^2\theta = \cos^2\theta-3\cos\theta+2 \Rightarrow 2\cos^2\theta-3\cos\theta+1 = 0\)
  \((2\cos\theta-1)(\cos\theta-1) = 0 \Rightarrow \cos\theta = \frac{1}{2} \Rightarrow \theta = 60^\circ\)
  \(\tan^2 30^\circ+\sin^2 30^\circ = \frac{1}{3}+\frac{1}{4} = \frac{7}{12}\)
  \(\tan 60^\circ+\sin 60^\circ = \sqrt{3}+\frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{2}\)
  मान \(= \frac{7}{12}\times \frac{2}{3\sqrt{3}} = \frac{7}{18\sqrt{3}} = \frac{7\sqrt{3}}{54}\)

\(\sec A – \cos A = ?\)

• \(\sin A\cdot\cot A\)
• \(\cot A\cdot\cos A\)
• \(\tan A\cdot\sin A\)
• \(a\cos A\cdot\sin A\)

Explanation:

• \(\sec A-\cos A = \frac{1}{\cos A}-\cos A = \frac{1-\cos^2 A}{\cos A} = \frac{\sin^2 A}{\cos A}\)
  \(= \sin A\times \frac{\sin A}{\cos A} = \tan A\cdot\sin A\)

यदि \(\sqrt{3}\tan\theta = 3\sin\theta\), तो \(\sin^2\theta – \cos^2\theta\) का मान क्या है?

• \(1/5\)
• \(1/4\)
• \(1/2\)
• \(1/3\)

Explanation:

• \(\sqrt{3}\tan\theta = 3\sin\theta \Rightarrow \frac{\sqrt{3}}{\cos\theta} = 3 \Rightarrow \cos\theta = \frac{1}{\sqrt{3}}\)
  \(\cos^2\theta = \frac{1}{3},\ \sin^2\theta = \frac{2}{3}\)
  \(\sin^2\theta-\cos^2\theta = \frac{2}{3}-\frac{1}{3} = \frac{1}{3}\)

\(\dfrac{1}{\tan\theta} + \tan\theta = ?\)

• \(\sec\theta\cdot\csc\theta\)
• \(1\)
• \(\dfrac{\csc\theta}{\sec\theta}\)
• \(\tan^2\theta\)

Explanation:

• \(\frac{1}{\tan\theta}+\tan\theta = \frac{1+\tan^2\theta}{\tan\theta} = \frac{\sec^2\theta}{\tan\theta}\)
  \(= \frac{1}{\cos^2\theta}\times \frac{\cos\theta}{\sin\theta} = \frac{1}{\sin\theta\cos\theta} = \sec\theta\cdot\csc\theta\)

निम्नलिखित विकल्पों में से गलत सूत्र का चयन करें।

• \(\csc^2\theta – \cot^2\theta = 1\)
• \(\sin^2\theta + \cos^2\theta = 1\)
• \(\sec^2\theta + \cos^2\theta = 1\)
• \(\sec^2\theta – \tan^2\theta = 1\)

Explanation:

• सही सूत्र: \(\csc^2\theta-\cot^2\theta = 1\), \(\sin^2\theta+\cos^2\theta = 1\), \(\sec^2\theta-\tan^2\theta = 1\)
  जाँच: \(\theta = 0^\circ\) पर \(\sec^2\theta+\cos^2\theta = 1+1 = 2 \neq 1\)
  अतः गलत सूत्र \(\sec^2\theta+\cos^2\theta = 1\) है

यदि \(a\cot\theta + b\csc\theta = p\) और \(b\cot\theta + a\csc\theta = q\) है, तो \(p^2 – q^2\) \(\_\_\_\_\) के बराबर है।

• \(b^2 – a^2\)
• \(a^2 – b^2\)
• \(b – a\)
• \(a^2 + b^2\)

Explanation:

• \(p^2 = a^2\cot^2\theta+b^2\csc^2\theta+2ab\cot\theta\csc\theta\)
  \(q^2 = b^2\cot^2\theta+a^2\csc^2\theta+2ab\cot\theta\csc\theta\)
  \(p^2-q^2 = a^2(\cot^2\theta-\csc^2\theta)+b^2(\csc^2\theta-\cot^2\theta) = -a^2+b^2 = b^2-a^2\)

\((\sin\theta + \csc\theta)^2 + (\cos\theta + \sec\theta)^2\)

• \(5 + \tan^2\theta + \cot^2\theta\)
• \(7 – \tan^2\theta + \cot^2\theta\)
• \(7 + \tan^2\theta + \cot^2\theta\)
• \(5 + \tan^2\theta – \cot^2\theta\)

Explanation:

• \((\sin\theta+\csc\theta)^2 = \sin^2\theta+2+\csc^2\theta\) और \((\cos\theta+\sec\theta)^2 = \cos^2\theta+2+\sec^2\theta\)
  जोड़ने पर \(= 1+4+\csc^2\theta+\sec^2\theta = 5+(1+\cot^2\theta)+(1+\tan^2\theta)\)
  \(= 7+\tan^2\theta+\cot^2\theta\)

\(\sin^2 60^\circ – \cos^2 45^\circ + \sec 60^\circ + \cos^2 40^\circ + \cos^2 50^\circ\) का मान है:

• \(\dfrac{13}{4}\)
• \(\dfrac{7}{2}\)
• \(\dfrac{11}{4}\)
• \(\dfrac{9}{14}\)

Explanation:

• \(\sin^2 60^\circ = \dfrac{3}{4},\ \cos^2 45^\circ = \dfrac{1}{2},\ \sec 60^\circ = 2\)
  \(\cos 50^\circ = \sin 40^\circ\), अतः \(\cos^2 40^\circ + \cos^2 50^\circ = 1\)
  मान \(= \dfrac{3}{4} – \dfrac{1}{2} + 2 + 1 = \dfrac{13}{4}\)

\(\dfrac{2\sin^2 30^\circ\tan 60^\circ – 3\cos^2 60^\circ\sec^2 30^\circ}{4\cot^2 45^\circ – \sec^2 60^\circ + \sin^2 60^\circ + \cos^2 90^\circ} = ?\)

• \(\dfrac{2(\sqrt{3}+2)}{3}\)
• \(\dfrac{1}{3}(\sqrt{3}-2)\)
• \(\dfrac{2(\sqrt{3}-2)}{3}\)
• \(\dfrac{1}{3}(\sqrt{3}+2)\)

Explanation:

• अंश \(= 2\times\dfrac{1}{4}\times\sqrt{3} – 3\times\dfrac{1}{4}\times\dfrac{4}{3} = \dfrac{\sqrt{3}}{2} – 1 = \dfrac{\sqrt{3} – 2}{2}\)
  हर \(= 4\times 1 – 4 + \dfrac{3}{4} + 0 = \dfrac{3}{4}\)
  मान \(= \dfrac{\sqrt{3}-2}{2}\times\dfrac{4}{3} = \dfrac{2(\sqrt{3}-2)}{3}\)

\(\sin^4 30^\circ + \cos^4 30^\circ – \sin 25^\circ\cos 65^\circ – \sin 65^\circ\cos 25^\circ = ?\)

• \(\dfrac{5}{8}\)
• \(-\dfrac{3}{8}\)
• \(\dfrac{13}{8}\)
• \(0\)

Explanation:

• \(\sin^4 30^\circ + \cos^4 30^\circ = \dfrac{1}{16} + \dfrac{9}{16} = \dfrac{5}{8}\)
  \(\sin 25^\circ\cos 65^\circ + \sin 65^\circ\cos 25^\circ = \sin(25^\circ + 65^\circ) = \sin 90^\circ = 1\)
  मान \(= \dfrac{5}{8} – 1 = -\dfrac{3}{8}\)

\(\sin^2 60^\circ\cos^2 45^\circ + 2\tan^2 60^\circ – \csc^2 30^\circ = ?\)

• \(-\dfrac{17}{24}\)
• \(-\dfrac{19}{8}\)
• \(\dfrac{17}{24}\)
• \(\dfrac{19}{8}\)

Explanation:

• \(\sin^2 60^\circ\cos^2 45^\circ = \dfrac{3}{4}\times\dfrac{1}{2} = \dfrac{3}{8}\)
  \(2\tan^2 60^\circ = 6,\ \operatorname{cosec}^2 30^\circ = 4\)
  मान \(= \dfrac{3}{8} + 6 – 4 = \dfrac{19}{8}\)

यदि \(2k\sin 30^\circ\cos 30^\circ\cot 60^\circ = \dfrac{\cot^2 30^\circ\sec 60^\circ\tan 45^\circ}{\csc^2 45^\circ\csc 30^\circ}\) है, तो \(k\) का मान ज्ञात कीजिए।

• \(\dfrac{3}{2}\)
• \(3\)
• \(1\)
• \(6\)

Explanation:

• बायाँ पक्ष: \(2k\times\dfrac{1}{2}\times\dfrac{\sqrt{3}}{2}\times\dfrac{1}{\sqrt{3}} = \dfrac{k}{2}\)
  दायाँ पक्ष: अंश \(= \cot^2 30^\circ\sec 60^\circ\tan 45^\circ = 3\times 2\times 1 = 6\), हर \(= \operatorname{cosec}^2 45^\circ\operatorname{cosec}30^\circ = 2\times 2 = 4\), अतः \(\dfrac{6}{4} = \dfrac{3}{2}\)
  \(\dfrac{k}{2} = \dfrac{3}{2} \Rightarrow k = 3\)

यदि \(A = 60^\circ\) है, तो \(\dfrac{[8\cos A + 7\sec A – \tan^2 A]}{10\sin\frac{A}{2}}\) का मान ज्ञात करें।

• \(5\)
• \(3\)
• \(15\)
• \(10\)

Explanation:

• \(A = 60^\circ\): \(8\cos 60^\circ = 4,\ 7\sec 60^\circ = 14,\ \tan^2 60^\circ = 3\)
  अंश \(= 4 + 14 – 3 = 15\)
  हर \(= 10\sin 30^\circ = 5\)
  मान \(= 3\)

यदि \(A = 10^\circ\) है, तो \(\dfrac{12\sin 3A + 5\cos(5A – 5^\circ)}{9\sin\frac{9A}{2} – 4\cos(5A + 10^\circ)}\) का मान ज्ञात करें।

• \(\dfrac{6\sqrt{2}+5}{(9-2\sqrt{2})}\)
• \(\dfrac{6\sqrt{2}-5}{(9-2\sqrt{2})}\)
• \(\dfrac{(9-2\sqrt{2})}{(6\sqrt{2}+5)}\)
• \(\dfrac{6\sqrt{2}+5}{(9+2\sqrt{2})}\)

Explanation:

• \(A = 10^\circ\): \(3A = 30^\circ,\ 5A – 5^\circ = 45^\circ,\ \dfrac{9A}{2} = 45^\circ,\ 5A + 10^\circ = 60^\circ\)
  अंश \(= 12\sin 30^\circ + 5\cos 45^\circ = 6 + \dfrac{5}{\sqrt{2}}\)
  हर \(= 9\sin 45^\circ – 4\cos 60^\circ = \dfrac{9}{\sqrt{2}} – 2\)
  अंश व हर को \(\sqrt{2}\) से गुणा करने पर मान \(= \dfrac{6\sqrt{2} + 5}{9 – 2\sqrt{2}}\)

यदि \(A = 60^\circ\) है, तो \(\dfrac{10\sin\frac{A}{2} + 8\cos A}{7\sin\frac{3A}{2} – 12\cos A}\) का मान ज्ञात करें।

• \(10\)
• \(12\)
• \(9\)
• \(7\)

Explanation:

• \(A = 60^\circ\): \(\dfrac{A}{2} = 30^\circ,\ \dfrac{3A}{2} = 90^\circ\)
  अंश \(= 10\sin 30^\circ + 8\cos 60^\circ = 5 + 4 = 9\)
  हर \(= 7\sin 90^\circ – 12\cos 60^\circ = 7 – 6 = 1\)
  मान \(= 9\)

यदि \(A = 30^\circ\) है, तो \(\dfrac{[8\sin A + 11\csc A – \cot^2 A]}{10\cos 2A}\) का मान ज्ञात करें।

• \(5\tfrac{1}{5}\)
• \(4\tfrac{3}{5}\)
• \(4\tfrac{2}{5}\)
• \(3\tfrac{4}{5}\)

Explanation:

• \(A = 30^\circ\): \(8\sin 30^\circ = 4,\ 11\operatorname{cosec}30^\circ = 22,\ \cot^2 30^\circ = 3\)
  अंश \(= 4 + 22 – 3 = 23\)
  हर \(= 10\cos 60^\circ = 5\)
  मान \(= \dfrac{23}{5} = 4\dfrac{3}{5}\)

\(\dfrac{\cos 45^\circ}{\sec 30^\circ + \csc 30^\circ} = ?\)

• \(\dfrac{3\sqrt{2}+\sqrt{6}}{8}\)
• \(\dfrac{\sqrt{6}}{2\sqrt{2}-2\sqrt{6}}\)
• \(\dfrac{3\sqrt{2}-\sqrt{6}}{8}\)
• \(\dfrac{\sqrt{3}}{2\sqrt{6}-2\sqrt{2}}\)

Explanation:

• \(\dfrac{\cos 45^\circ}{\sec 30^\circ + \operatorname{cosec}30^\circ} = \dfrac{\frac{1}{\sqrt{2}}}{\frac{2}{\sqrt{3}} + 2} = \dfrac{\sqrt{3}}{2\sqrt{2}(1 + \sqrt{3})}\)
  \((\sqrt{3} – 1)\) से गुणा-भाग करने पर \(= \dfrac{\sqrt{3}(\sqrt{3} – 1)}{2\sqrt{2}\times 2} = \dfrac{3 – \sqrt{3}}{4\sqrt{2}}\)
  \(= \dfrac{(3 – \sqrt{3})\sqrt{2}}{8} = \dfrac{3\sqrt{2} – \sqrt{6}}{8}\)

\(\dfrac{\sin 30^\circ\sin 40^\circ\sin 50^\circ\sin 60^\circ}{\cos 30^\circ\cos 40^\circ\cos 50^\circ\cos 60^\circ} = ?\)

• \(\dfrac{1}{\sqrt{3}}\)
• \(\sqrt{3}\)
• \(1\)
• \(\dfrac{1}{\sqrt{2}}\)

Explanation:

• \(\sin 40^\circ = \cos 50^\circ\) और \(\sin 50^\circ = \cos 40^\circ\)
  \(\dfrac{\sin 30^\circ}{\cos 60^\circ} = 1,\ \dfrac{\sin 60^\circ}{\cos 30^\circ} = 1,\ \dfrac{\sin 40^\circ}{\cos 50^\circ} = 1,\ \dfrac{\sin 50^\circ}{\cos 40^\circ} = 1\)
  गुणनफल \(= 1\)

\((\sin 30^\circ\cos 60^\circ – \cos 30^\circ\sin 60^\circ) = ?\)

• \(-\cos 30^\circ\)
• \(-\sin 30^\circ\)
• \(\cos 30^\circ\)
• \(\sin 30^\circ\)

Explanation:

• \(\sin A\cos B – \cos A\sin B = \sin(A – B)\)
  \(\sin(30^\circ – 60^\circ) = \sin(-30^\circ) = -\sin 30^\circ\)

\(16\cos^3\dfrac{\pi}{6} – 12\cos\dfrac{\pi}{6} = \_\_\_\_\_\_\)

• \(0\)
• \(2\)
• \(1\)
• \(-1\)

Explanation:

• \(4\cos^3\theta – 3\cos\theta = \cos 3\theta\), अतः \(16\cos^3\theta – 12\cos\theta = 4\cos 3\theta\)
  \(\theta = \dfrac{\pi}{6} \Rightarrow 3\theta = \dfrac{\pi}{2}\)
  \(4\cos\dfrac{\pi}{2} = 0\)
  (सीधी गणना: \(16\times\dfrac{3\sqrt{3}}{8} – 12\times\dfrac{\sqrt{3}}{2} = 6\sqrt{3} – 6\sqrt{3} = 0\))

\(\cos^3 60^\circ – \cos^3 240^\circ – \cos^3 360^\circ = \_\_\_\_\_\_\_\_\)

• \(-\dfrac{3}{4}\)
• \(-\dfrac{7}{5}\)
• \(-\dfrac{3}{5}\)
• \(-\dfrac{9}{5}\)

Explanation:

• \(\cos 60^\circ = \dfrac{1}{2} \Rightarrow \cos^3 60^\circ = \dfrac{1}{8}\)
  \(\cos 240^\circ = -\dfrac{1}{2} \Rightarrow \cos^3 240^\circ = -\dfrac{1}{8}\)
  \(\cos 360^\circ = 1\)
  मान \(= \dfrac{1}{8} + \dfrac{1}{8} – 1 = -\dfrac{3}{4}\)

यदि \(\sin(A + B) = \dfrac{\sqrt{3}}{2}\) और \(\tan(A – B) = \dfrac{1}{\sqrt{3}}\) तो \((2A + 3B)\) बराबर है:

• \(120^\circ\)
• \(135^\circ\)
• \(130^\circ\)
• \(125^\circ\)

Explanation:

• \(\sin(A+B) = \dfrac{\sqrt{3}}{2} \Rightarrow A + B = 60^\circ\)
  \(\tan(A-B) = \dfrac{1}{\sqrt{3}} \Rightarrow A – B = 30^\circ\)
  \(A = 45^\circ,\ B = 15^\circ\)
  \(2A + 3B = 90^\circ + 45^\circ = 135^\circ\)

\(\tan^2 48^\circ – \csc^2 42^\circ + \csc(67^\circ + \theta) – \sec(23^\circ – \theta)\) का मान ज्ञात कीजिए।

• \(-1\)
• \(0\)
• \(1\)
• \(-2\)

Explanation:

• \(\operatorname{cosec}42^\circ = \sec 48^\circ\), अतः \(\tan^2 48^\circ – \operatorname{cosec}^2 42^\circ = \tan^2 48^\circ – \sec^2 48^\circ = -1\)
  \((67^\circ + \theta) + (23^\circ – \theta) = 90^\circ\), अतः \(\operatorname{cosec}(67^\circ+\theta) – \sec(23^\circ-\theta) = 0\)
  मान \(= -1\)

\((\sin^4 45^\circ + \cos^4 60^\circ) + (\tan^4 45^\circ + \cot^4 45^\circ) = ?\)

• \(\dfrac{33}{16}\)
• \(\dfrac{37}{16}\)
• \(\dfrac{35}{16}\)
• \(\dfrac{39}{16}\)

Explanation:

• \(\sin^4 45^\circ = \dfrac{1}{4},\ \cos^4 60^\circ = \dfrac{1}{16},\ \tan^4 45^\circ = 1,\ \cot^4 45^\circ = 1\)
  मान \(= \dfrac{1}{4} + \dfrac{1}{16} + 1 + 1 = \dfrac{4 + 1 + 32}{16} = \dfrac{37}{16}\)

\(\cos\left(-\dfrac{7\pi}{2}\right) = ?\)

• \(\dfrac{1}{2}\)
• \(1\)
• \(-1\)
• \(0\)

Explanation:

• \(\cos(-x) = \cos x \Rightarrow \cos\left(-\dfrac{7\pi}{2}\right) = \cos\dfrac{7\pi}{2}\)
  \(\dfrac{7\pi}{2} = 3\pi + \dfrac{\pi}{2}\), अर्थात \(\dfrac{\pi}{2}\) का विषम गुणज, इसलिए \(\cos\dfrac{7\pi}{2} = 0\)
  नोट: सही मान \(0\) (विकल्प 4) आता है; कुंजी में \(-1\) दिया है, जो सही नहीं है। कृपया कुंजी जाँच लें।

व्यंजक \(\cos^2 45^\circ + \cos^2 135^\circ + \cos^2 225^\circ + \cos^2 315^\circ\) का मान है-

• \(2\)
• \(\dfrac{1}{2}\)
• \(\dfrac{3}{2}\)
• \(1\)

Explanation:

• \(\cos^2 135^\circ = \cos^2(180^\circ – 45^\circ) = \dfrac{1}{2}\)
  \(\cos^2 225^\circ = \cos^2(180^\circ + 45^\circ) = \dfrac{1}{2}\)
  \(\cos^2 315^\circ = \cos^2(360^\circ – 45^\circ) = \dfrac{1}{2}\)
  योग \(= 4\times\dfrac{1}{2} = 2\)

यदि \(\cos 48^\circ = \dfrac{m}{n}\) है, तो \(\sec 48^\circ – \cot 42^\circ\) \(\_\_\_\_\_\) के बराबर है।

• \(\dfrac{m – \sqrt{n^2 – m^2}}{m}\)
• \(\dfrac{m – \sqrt{n^2 – m^2}}{n}\)
• \(\dfrac{n – \sqrt{n^2 – m^2}}{n}\)
• \(\dfrac{n – \sqrt{n^2 – m^2}}{m}\)

Explanation:

• \(\cos 48^\circ = \dfrac{m}{n} \Rightarrow \sec 48^\circ = \dfrac{n}{m},\ \sin 48^\circ = \dfrac{\sqrt{n^2 – m^2}}{n}\)
  \(\cot 42^\circ = \tan 48^\circ = \dfrac{\sin 48^\circ}{\cos 48^\circ} = \dfrac{\sqrt{n^2 – m^2}}{m}\)
  \(\sec 48^\circ – \cot 42^\circ = \dfrac{n – \sqrt{n^2 – m^2}}{m}\)

\(\dfrac{3\tan^2 60^\circ + \sec^2 30^\circ – \sin^2 45^\circ}{(\cos 15^\circ + \sin 75^\circ)(\sec 15^\circ + \csc 75^\circ)}\) का मान ज्ञात करें।

• \(\dfrac{5}{12}\)
• \(\dfrac{65}{24}\)
• \(\dfrac{13}{12}\)
• \(\dfrac{59}{24}\)

Explanation:

• अंश \(= 3\times 3 + \dfrac{4}{3} – \dfrac{1}{2} = \dfrac{54 + 8 – 3}{6} = \dfrac{59}{6}\)
  हर: \(\sin 75^\circ = \cos 15^\circ\) और \(\operatorname{cosec}75^\circ = \sec 15^\circ\), अतः \((2\cos 15^\circ)(2\sec 15^\circ) = 4\)
  मान \(= \dfrac{59}{6\times 4} = \dfrac{59}{24}\)

\(\cot 75^\circ = 2 – \sqrt{3}\) है। \(\cot 15^\circ\) का मान ज्ञात कीजिए।

• \(2 – \sqrt{3}\)
• \(2 + \sqrt{3}\)
• \(\sqrt{3} + 1\)
• \(\sqrt{3} – 1\)

Explanation:

• \(\cot 15^\circ = \tan 75^\circ = \dfrac{1}{\cot 75^\circ} = \dfrac{1}{2 – \sqrt{3}}\)
  \(= \dfrac{2 + \sqrt{3}}{(2 – \sqrt{3})(2 + \sqrt{3})} = 2 + \sqrt{3}\)

\(\left[\dfrac{\csc 31^\circ – \sec 59^\circ}{\sin 18^\circ + \cos 72^\circ} \times \dfrac{\tan 26^\circ}{\cot 64^\circ}\right] + \left[\dfrac{\tan 30^\circ + \tan 15^\circ}{1 – \tan 30^\circ\tan 15^\circ}\right] = ?\)

• \(0\)
• \(-1\)
• \(1\)
• \(2\)

Explanation:

• \(\operatorname{cosec}31^\circ = \sec 59^\circ\), अतः पहले कोष्ठक का अंश \(0\) है और पहला भाग \(= 0\)
  दूसरा भाग: \(\dfrac{\tan 30^\circ + \tan 15^\circ}{1 – \tan 30^\circ\tan 15^\circ} = \tan(30^\circ + 15^\circ) = \tan 45^\circ = 1\)
  मान \(= 0 + 1 = 1\)

\(\dfrac{\csc^2 30^\circ\sin^2 45^\circ + \sec^2 60^\circ}{\tan 60^\circ\csc^2 45^\circ – \sec^2 60^\circ\tan 45^\circ} = ?\)

• \(3(2 + \sqrt{3})\)
• \(2(\sqrt{3} – 2)\)
• \(-2\sqrt{3} – 2\)
• \(-3(2 + \sqrt{3})\)

Explanation:

• अंश \(= \operatorname{cosec}^2 30^\circ\sin^2 45^\circ + \sec^2 60^\circ = 4\times\dfrac{1}{2} + 4 = 6\)
  हर \(= \tan 60^\circ\operatorname{cosec}^2 45^\circ – \sec^2 60^\circ\tan 45^\circ = 2\sqrt{3} – 4\)
  मान \(= \dfrac{6}{2\sqrt{3} – 4} = \dfrac{3}{\sqrt{3} – 2} = \dfrac{3(\sqrt{3} + 2)}{3 – 4} = -3(2 + \sqrt{3})\)

\(\dfrac{4\tan^2 30^\circ + \sin^2 30^\circ\cos^2 45^\circ + \sec^2 48^\circ – \cot^2 42^\circ}{\cos 37^\circ\sin 53^\circ + \sin 37^\circ\cos 53^\circ + \tan 18^\circ\tan 72^\circ}\)

• \(\dfrac{35}{48}\)
• \(\dfrac{59}{48}\)
• \(\dfrac{49}{24}\)
• \(\dfrac{35}{24}\)

Explanation:

• अंश: \(4\tan^2 30^\circ = \dfrac{4}{3},\ \sin^2 30^\circ\cos^2 45^\circ = \dfrac{1}{8}\)
  \(\sec 48^\circ = \operatorname{cosec}42^\circ\), \(\cot 42^\circ = \tan 48^\circ\): \(\sec^2 48^\circ – \tan^2 48^\circ = 1\)
  अंश \(= \dfrac{4}{3} + \dfrac{1}{8} + 1 = \dfrac{59}{24}\)
  हर: \(\cos 37^\circ\sin 53^\circ + \sin 37^\circ\cos 53^\circ = \sin 90^\circ = 1\) और \(\tan 18^\circ\tan 72^\circ = 1\), अतः हर \(= 2\)
  मान \(= \dfrac{59}{48}\)

यदि \(0^\circ < \theta < 90^\circ\) और \(\cos^2\theta = 3(\cot^2\theta – \cos^2\theta)\) है, तो \(\left(\dfrac{1}{2}\sec\theta + \sin\theta\right)^{-1}\) का मान होगा:

• \(\sqrt{3} + 2\)
• \(2(2 – \sqrt{3})\)
• \(2(\sqrt{3} – 1)\)
• \(\sqrt{3} + 1\)

Explanation:

• \(\cot^2\theta – \cos^2\theta = \dfrac{\cos^4\theta}{\sin^2\theta}\)
  \(\cos^2\theta = \dfrac{3\cos^4\theta}{\sin^2\theta} \Rightarrow \tan^2\theta = 3 \Rightarrow \theta = 60^\circ\)
  \(\left(\dfrac{1}{2}\sec 60^\circ + \sin 60^\circ\right)^{-1} = \left(1 + \dfrac{\sqrt{3}}{2}\right)^{-1} = \dfrac{2}{2 + \sqrt{3}} = 2(2 – \sqrt{3})\)

\(\left(\dfrac{\sin\theta – 2\sin^3\theta}{2\cos^3\theta – \cos\theta}\right)^2 + 1,\ \theta \ne 45^\circ\) बराबर है:

• \(\csc^2\theta\)
• \(\sec^2\theta\)
• \(\cot^2\theta\)
• \(2\tan^2\theta\)

Explanation:

• \(\sin\theta – 2\sin^3\theta = \sin\theta(1 – 2\sin^2\theta) = \sin\theta\cos 2\theta\)
  \(2\cos^3\theta – \cos\theta = \cos\theta(2\cos^2\theta – 1) = \cos\theta\cos 2\theta\)
  अनुपात \(= \tan\theta\) (\(\theta \ne 45^\circ\), अतः \(\cos 2\theta \ne 0\))
  \(\tan^2\theta + 1 = \sec^2\theta\)

\(\sqrt{\sec^2\theta + \csc^2\theta} \times \sqrt{\tan^2\theta – \sin^2\theta} = ?\)

• \(\csc\theta\cdot\sec^2\theta\)
• \(\sin\theta\cdot\sec^2\theta\)
• \(\sin\theta\cdot\cos^2\theta\)
• \(\csc\theta\cdot\cos^2\theta\)

Explanation:

• \(\sec^2\theta + \operatorname{cosec}^2\theta = \dfrac{1}{\cos^2\theta} + \dfrac{1}{\sin^2\theta} = \dfrac{1}{\sin^2\theta\cos^2\theta}\), अतः पहला मूल \(= \dfrac{1}{\sin\theta\cos\theta}\)
  \(\tan^2\theta – \sin^2\theta = \tan^2\theta(1 – \cos^2\theta) = \dfrac{\sin^4\theta}{\cos^2\theta}\), अतः दूसरा मूल \(= \dfrac{\sin^2\theta}{\cos\theta}\)
  गुणनफल \(= \dfrac{\sin\theta}{\cos^2\theta} = \sin\theta\sec^2\theta\)

माना \(a = \dfrac{2\sin x}{1 + \sin x + \cos x}\) और \(b = \dfrac{c}{1 + \sin x}\) तो \(a = b\) है, तो \(c = ?\)

• \(1 – \sin x\cdot\cos x\)
• \(1 + \sin x – \cos x\)
• \(1 + \sin x\cdot\cos x\)
• \(1 + \cos x – \sin x\)

Explanation:

• \(a = b \Rightarrow c = \dfrac{2\sin x(1 + \sin x)}{1 + \sin x + \cos x}\)
  अंश व हर को \((1 + \sin x – \cos x)\) से गुणा करने पर हर \(= (1 + \sin x)^2 – \cos^2 x = 2\sin x + 2\sin^2 x = 2\sin x(1 + \sin x)\)
  अतः \(c = 1 + \sin x – \cos x\)

यदि \((1 + \tan^2\theta) + (1 + (\tan^2\theta)^{-1}) = k\), तो \(\sqrt{k} = ?\)

• \(\csc\theta\cdot\sec\theta\)
• \(\csc\theta\cdot\cos\theta\)
• \(\sin\theta\cdot\cos\theta\)
• \(\sin\theta\cdot\sec\theta\)

Explanation:

• \((1 + \tan^2\theta) + (1 + \cot^2\theta) = \sec^2\theta + \operatorname{cosec}^2\theta = \dfrac{1}{\sin^2\theta\cos^2\theta} = k\)
  \(\sqrt{k} = \dfrac{1}{\sin\theta\cos\theta} = \operatorname{cosec}\theta\sec\theta\)

यदि \(\sin\theta = \dfrac{a}{\sqrt{a^2 + b^2}},\ 0^\circ < \theta < 90^\circ\), तो \(\sec\theta + \tan\theta\) का मान है:

• \(\dfrac{\sqrt{a^2+b^2} + a}{b}\)
• \(\dfrac{\sqrt{a^2+b^2} + b}{2a}\)
• \(\dfrac{\sqrt{a^2+b^2} + a}{2b}\)
• \(\dfrac{\sqrt{a^2+b^2} + b}{a}\)

Explanation:

• \(\sin\theta = \dfrac{a}{\sqrt{a^2 + b^2}} \Rightarrow \cos\theta = \dfrac{b}{\sqrt{a^2 + b^2}}\)
  \(\sec\theta + \tan\theta = \dfrac{1 + \sin\theta}{\cos\theta} = \dfrac{\sqrt{a^2+b^2} + a}{b}\)

यदि \(\left(\dfrac{\tan\theta – \sec\theta + 1}{\tan\theta + \sec\theta – 1}\right)\sec\theta = \dfrac{1}{k}\), तो \(k = \_\_\_\_\_\_\).

• \(1 + \sin\theta\)
• \(1 – \cos\theta\)
• \(1 + \cos\theta\)
• \(1 – \sin\theta\)

Explanation:

• अंश व हर को \(\cos\theta\) से गुणा करने पर \(\dfrac{\sin\theta – 1 + \cos\theta}{\sin\theta + 1 – \cos\theta}\)
  अर्ध-कोण से: अंश \(= 2\sin\frac{\theta}{2}\left(\cos\frac{\theta}{2} – \sin\frac{\theta}{2}\right)\), हर \(= 2\sin\frac{\theta}{2}\left(\cos\frac{\theta}{2} + \sin\frac{\theta}{2}\right)\)
  इसलिए अनुपात \(= \dfrac{\cos\frac{\theta}{2} – \sin\frac{\theta}{2}}{\cos\frac{\theta}{2} + \sin\frac{\theta}{2}}\)
  \(\sec\theta\) से गुणा करने पर \(\dfrac{1}{(\cos\frac{\theta}{2} + \sin\frac{\theta}{2})^2} = \dfrac{1}{1 + \sin\theta} = \dfrac{1}{k}\)
  \(k = 1 + \sin\theta\)

यदि \(\dfrac{\tan\theta}{1 – \cot\theta} + \dfrac{\cot\theta}{1 – \tan\theta} = 1 + K\), तो \(K = \_\_\_\_\).

• \(\cot\theta + \sec\theta\)
• \(\tan\theta\cdot\csc\theta\)
• \(\tan\theta + \sec\theta\)
• \(\csc\theta\cdot\sec\theta\)

Explanation:

• \(\dfrac{\tan\theta}{1 – \cot\theta} = \dfrac{\tan^2\theta}{\tan\theta – 1}\) और \(\dfrac{\cot\theta}{1 – \tan\theta} = -\dfrac{\cot\theta}{\tan\theta – 1}\)
  योग \(= \dfrac{\tan^2\theta – \cot\theta}{\tan\theta – 1} = \dfrac{t^3 – 1}{t(t – 1)} = \dfrac{t^2 + t + 1}{t}\) (जहाँ \(t = \tan\theta\))
  \(= 1 + \tan\theta + \cot\theta = 1 + \sec\theta\operatorname{cosec}\theta\)
  अतः \(K = \operatorname{cosec}\theta\sec\theta\)

यदि \(x = 4\cos A + 5\sin A\) और \(y = 4\sin A – 5\cos A\) हो, तो \(x^2 + y^2\) का मान ज्ञात कीजिए।

• \(25\)
• \(16\)
• \(0\)
• \(41\)

Explanation:

• \(x^2 + y^2 = (4\cos A + 5\sin A)^2 + (4\sin A – 5\cos A)^2\)
  \(= 16\cos^2 A + 40\sin A\cos A + 25\sin^2 A + 16\sin^2 A – 40\sin A\cos A + 25\cos^2 A\)
  \(= 16 + 25 = 41\)

\(\dfrac{\sec^6\theta – \tan^6\theta – 3\sec^2\theta\tan^2\theta + 1}{\cos^4\theta – \sin^4\theta + 2\sin^2\theta + 2} = ?\)

• \(\dfrac{3}{4}\)
• \(\dfrac{2}{3}\)
• \(\dfrac{1}{2}\)
• \(1\)

Explanation:

• \(\sec^6\theta – \tan^6\theta = (\sec^2\theta – \tan^2\theta)(\sec^4\theta + \sec^2\theta\tan^2\theta + \tan^4\theta) = \sec^4\theta + \sec^2\theta\tan^2\theta + \tan^4\theta\)
  अंश \(= \sec^4\theta – 2\sec^2\theta\tan^2\theta + \tan^4\theta + 1 = (\sec^2\theta – \tan^2\theta)^2 + 1 = 2\)
  हर: \(\cos^4\theta – \sin^4\theta = \cos^2\theta – \sin^2\theta\), अतः हर \(= \cos^2\theta + \sin^2\theta + 2 = 3\)
  मान \(= \dfrac{2}{3}\)

यदि \(x\cos A – y\sin A = 1\) और \(x\sin A + y\cos A = 4\) है, तो \(17x^2 + 17y^2\) का मान बताइए।

• \(0\)
• \(7\)
• \(49\)
• \(289\)

Explanation:

• \((x\cos A – y\sin A)^2 + (x\sin A + y\cos A)^2 = x^2 + y^2\)
  \(x^2 + y^2 = 1^2 + 4^2 = 17\)
  \(17(x^2 + y^2) = 17\times 17 = 289\)

यदि \(2\sin A + \csc A = 2\sqrt{2},\ 0^\circ < A < 90^\circ\) है, तो \(2(\sin^4 A + \cos^4 A)\) का मान बताइए।

• \(2\)
• \(1\)
• \(4\)
• \(0\)

Explanation:

• \(2\sin A + \dfrac{1}{\sin A} = 2\sqrt{2} \Rightarrow 2\sin^2 A – 2\sqrt{2}\sin A + 1 = 0\)
  \((\sqrt{2}\sin A – 1)^2 = 0 \Rightarrow \sin A = \dfrac{1}{\sqrt{2}},\ A = 45^\circ\)
  \(2(\sin^4 45^\circ + \cos^4 45^\circ) = 2\left(\dfrac{1}{4} + \dfrac{1}{4}\right) = 1\)

\(\dfrac{\tan(45^\circ – \alpha)}{\cot(45^\circ + \alpha)} – \dfrac{(\cos 19^\circ + \sin 71^\circ)(\sec 19^\circ + \csc 71^\circ)}{\tan 12^\circ\tan 24^\circ\tan 66^\circ\tan 78^\circ}\) का मान बताइए।

• \(-3\)
• \(2\)
• \(-2\)
• \(3\)

Explanation:

• \(\cot(45^\circ + \alpha) = \tan(45^\circ – \alpha)\), अतः पहला भाग \(= 1\)
  \((\cos 19^\circ + \sin 71^\circ)(\sec 19^\circ + \operatorname{cosec}71^\circ) = (2\cos 19^\circ)(2\sec 19^\circ) = 4\)
  हर: \(\tan 12^\circ\tan 78^\circ = 1\) और \(\tan 24^\circ\tan 66^\circ = 1\), अतः हर \(= 1\)
  मान \(= 1 – 4 = -3\)

\(\dfrac{\cot^3\theta}{\csc^2\theta} + \dfrac{\tan^3\theta}{\sec^2\theta} + 2\sin\theta\cos\theta = ?\)

• \(\csc\theta\sec\theta\)
• \(\csc^2\theta\sec^2\theta\)
• \(\sin\theta\cos\theta\)
• \(\sin^2\theta\cos\theta\)

Explanation:

• \(\dfrac{\cot^3\theta}{\operatorname{cosec}^2\theta} = \dfrac{\cos^3\theta}{\sin^3\theta}\times\sin^2\theta = \dfrac{\cos^3\theta}{\sin\theta}\) और \(\dfrac{\tan^3\theta}{\sec^2\theta} = \dfrac{\sin^3\theta}{\cos\theta}\)
  योग \(= \dfrac{\cos^4\theta + \sin^4\theta}{\sin\theta\cos\theta}\)
  \(2\sin\theta\cos\theta = \dfrac{2\sin^2\theta\cos^2\theta}{\sin\theta\cos\theta}\) जोड़ने पर अंश \(= (\sin^2\theta + \cos^2\theta)^2 = 1\)
  मान \(= \dfrac{1}{\sin\theta\cos\theta} = \operatorname{cosec}\theta\sec\theta\)

यदि \(\sin\left(\dfrac{2A+B}{2}\right) = \cos\left(\dfrac{2A-B}{2}\right) = \dfrac{\sqrt{3}}{2}\), \(0^\circ < \dfrac{2A+B}{2} < 90^\circ\) और \(0^\circ < \dfrac{2A-B}{2} < 90^\circ\) है, तो \(\sin[3(A-B)]\) का मान ज्ञात करें।

• \(1\)
• \(\dfrac{1}{\sqrt{2}}\)
• \(\dfrac{1}{2}\)
• \(\dfrac{\sqrt{3}}{2}\)

Explanation:

• \(\sin\dfrac{2A+B}{2} = \dfrac{\sqrt{3}}{2} \Rightarrow \dfrac{2A+B}{2} = 60^\circ \Rightarrow 2A + B = 120^\circ\)
  \(\cos\dfrac{2A-B}{2} = \dfrac{\sqrt{3}}{2} \Rightarrow \dfrac{2A-B}{2} = 30^\circ \Rightarrow 2A – B = 60^\circ\)
  जोड़ने-घटाने पर \(A = 45^\circ,\ B = 30^\circ\)
  \(\sin[3(A – B)] = \sin 45^\circ = \dfrac{1}{\sqrt{2}}\)

\(\left(\sqrt{\sec^2\theta + \csc^2\theta}\right)\left(\dfrac{\sin\theta(1+\cos\theta)}{1 + \cos\theta – \sin^2\theta}\right)\) \(\_\_\_\_\_\) के बराबर है, जहां \(0^\circ < \theta < 90^\circ\) है।

• \(\cot\theta\)
• \(\csc^2\theta\)
• \(\sec^2\theta\)
• \(\tan\theta\)

Explanation:

• \(\sqrt{\sec^2\theta + \operatorname{cosec}^2\theta} = \dfrac{1}{\sin\theta\cos\theta}\)
  \(1 + \cos\theta – \sin^2\theta = \cos\theta + \cos^2\theta = \cos\theta(1 + \cos\theta)\)
  दूसरा भाग \(= \dfrac{\sin\theta(1 + \cos\theta)}{\cos\theta(1 + \cos\theta)} = \tan\theta\)
  गुणनफल \(= \dfrac{1}{\sin\theta\cos\theta}\times\dfrac{\sin\theta}{\cos\theta} = \sec^2\theta\)

\(\dfrac{\sqrt{2}\tan(60^\circ – \theta)\tan(30^\circ + \theta)}{\sin^2(45^\circ + \theta) + \sin^2(45^\circ – \theta)} = ?\)

• \(\dfrac{1}{\sqrt{2}}\)
• \(1\)
• \(2\)
• \(\sqrt{2}\)

Explanation:

• \((60^\circ – \theta) + (30^\circ + \theta) = 90^\circ\), अतः \(\tan(60^\circ-\theta)\tan(30^\circ+\theta) = 1\)
  \(\sin(45^\circ – \theta) = \cos(45^\circ + \theta)\), अतः \(\sin^2(45^\circ+\theta) + \sin^2(45^\circ-\theta) = 1\)
  मान \(= \sqrt{2}\)

\((\sin 37^\circ\cos 53^\circ + \cos 37^\circ\sin 53^\circ) – \dfrac{4\cos^2 37^\circ – 7 + 4\cos^2 53^\circ}{\tan^2 47^\circ + 4 – \csc^2 43^\circ} = ?\)

• \(1\)
• \(-2\)
• \(0\)
• \(2\)

Explanation:

• \(\sin 37^\circ\cos 53^\circ + \cos 37^\circ\sin 53^\circ = \sin 90^\circ = 1\)
  \(\cos 53^\circ = \sin 37^\circ\): \(4(\cos^2 37^\circ + \sin^2 37^\circ) – 7 = -3\)
  \(\tan^2 47^\circ – \operatorname{cosec}^2 43^\circ = \tan^2 47^\circ – \sec^2 47^\circ = -1\), अतः हर \(= -1 + 4 = 3\)
  दूसरा भाग \(= \dfrac{-3}{3} = -1\)
  मान \(= 1 – (-1) = 2\)

यदि \(\sin^6\theta + \cos^6\theta = \dfrac{1}{3},\ 0^\circ < \theta < 90^\circ\) तो \(\sin\theta\cos\theta\) का मान ज्ञात करें।

• \(\dfrac{\sqrt{6}}{6}\)
• \(\dfrac{\sqrt{2}}{3}\)
• \(\dfrac{\sqrt{2}}{\sqrt{3}}\)
• \(\dfrac{2}{9}\)

Explanation:

• \(\sin^6\theta + \cos^6\theta = 1 – 3\sin^2\theta\cos^2\theta = \dfrac{1}{3}\)
  \(\sin^2\theta\cos^2\theta = \dfrac{2}{9} \Rightarrow \sin\theta\cos\theta = \dfrac{\sqrt{2}}{3}\)

\(\sec^4\theta(1 – \sin^4\theta) – 2\tan^2\theta = ?\)

• \(1\)
• \(0\)
• \(-1\)
• \(\dfrac{1}{2}\)

Explanation:

• \(1 – \sin^4\theta = (1 – \sin^2\theta)(1 + \sin^2\theta) = \cos^2\theta(1 + \sin^2\theta)\)
  \(\sec^4\theta\times\cos^2\theta(1 + \sin^2\theta) = \sec^2\theta(1 + \sin^2\theta) = \sec^2\theta + \tan^2\theta\)
  मान \(= \sec^2\theta + \tan^2\theta – 2\tan^2\theta = \sec^2\theta – \tan^2\theta = 1\)

यदि \(0^\circ < \theta < 90^\circ\) है, तो \(\sqrt{\dfrac{\sec^2\theta + \csc^2\theta}{\tan^2\theta – \sin^2\theta}}\) का मान ज्ञात करें।

• \(\sec^3\theta\)
• \(\sin^2\theta\)
• \(\csc^3\theta\)
• \(\sec^2\theta\)

Explanation:

• \(\sec^2\theta + \operatorname{cosec}^2\theta = \dfrac{1}{\sin^2\theta\cos^2\theta}\)
  \(\tan^2\theta – \sin^2\theta = \dfrac{\sin^4\theta}{\cos^2\theta}\)
  अनुपात \(= \dfrac{1}{\sin^2\theta\cos^2\theta}\times\dfrac{\cos^2\theta}{\sin^4\theta} = \dfrac{1}{\sin^6\theta}\)
  वर्गमूल \(= \dfrac{1}{\sin^3\theta} = \operatorname{cosec}^3\theta\)

यदि \(\dfrac{\sin^2\theta}{\tan^2\theta – \sin^2\theta} = 5\) है, \(\theta\) न्यून कोण है, तो \(\dfrac{24\sin^2\theta – 15\sec^2\theta}{6\csc^2\theta – 7\cot^2\theta}\) का मान ज्ञात करें।

• \(2\)
• \(-14\)
• \(14\)
• \(-2\)

Explanation:

• \(\tan^2\theta – \sin^2\theta = \dfrac{\sin^4\theta}{\cos^2\theta}\), अतः \(\dfrac{\sin^2\theta}{\tan^2\theta – \sin^2\theta} = \dfrac{\cos^2\theta}{\sin^2\theta} = \cot^2\theta = 5\)
  \(\sin^2\theta = \dfrac{1}{1 + \cot^2\theta} = \dfrac{1}{6},\ \sec^2\theta = 1 + \tan^2\theta = \dfrac{6}{5},\ \operatorname{cosec}^2\theta = 6\)
  अंश \(= 24\times\dfrac{1}{6} – 15\times\dfrac{6}{5} = 4 – 18 = -14\)
  हर \(= 6\times 6 – 7\times 5 = 1\)
  मान \(= -14\)

यदि \(0^\circ < \theta < 90^\circ\) है, तो \(1 + \sqrt{\dfrac{\cot\theta + \cos\theta}{\cot\theta – \cos\theta}}\) का मान ज्ञात कीजिए।

• \(1 – \sec\theta + \tan\theta\)
• \(1 – \sec\theta – \tan\theta\)
• \(1 + \sec\theta – \tan\theta\)
• \(1 + \sec\theta + \tan\theta\)

Explanation:

• \(\cot\theta = \dfrac{\cos\theta}{\sin\theta}\); अंश व हर को \(\cos\theta\) से भाग देने पर \(\dfrac{\cot\theta + \cos\theta}{\cot\theta – \cos\theta} = \dfrac{1 + \sin\theta}{1 – \sin\theta}\)
  \(\sqrt{\dfrac{1 + \sin\theta}{1 – \sin\theta}} = \dfrac{1 + \sin\theta}{\cos\theta} = \sec\theta + \tan\theta\)
  मान \(= 1 + \sec\theta + \tan\theta\)

व्यंजक \((\cos^6\theta + \sin^6\theta – 1)(\tan^2\theta + \cot^2\theta + 2) + 1\) का मान ज्ञात कीजिए।

• \(1\)
• \(-2\)
• \(0\)
• \(-1\)

Explanation:

• \(\cos^6\theta + \sin^6\theta – 1 = -3\sin^2\theta\cos^2\theta\)
  \(\tan^2\theta + \cot^2\theta + 2 = (\tan\theta + \cot\theta)^2 = \dfrac{1}{\sin^2\theta\cos^2\theta}\)
  गुणनफल \(= -3\), अतः मान \(= -3 + 1 = -2\)

\(2 – \sqrt{\dfrac{\cot\theta + \cos\theta}{\cot\theta – \cos\theta}}\) का मान बताइए, जबकि \(0^\circ < \theta < 90^\circ\) है।

• \(2 + \sec\theta + \tan\theta\)
• \(2 – \sec\theta + \tan\theta\)
• \(2 – \sec\theta – \tan\theta\)
• \(2 + \sec\theta – \tan\theta\)

Explanation:

• पिछले प्रश्न की भाँति \(\sqrt{\dfrac{\cot\theta + \cos\theta}{\cot\theta – \cos\theta}} = \sec\theta + \tan\theta\)
  मान \(= 2 – \sec\theta – \tan\theta\)

निम्नलिखित व्यंजक का मान बताइए: \(\dfrac{\cos A}{1 – \tan A} + \dfrac{\sin A}{1 – \cot A} – \sin A\)

• \(1 + \cos A\)
• \((1 + \sin A)\cos A\)
• \(1 + \sin A\)
• \(\cos A\)

Explanation:

• \(\dfrac{\cos A}{1 – \tan A} = \dfrac{\cos^2 A}{\cos A – \sin A}\) और \(\dfrac{\sin A}{1 – \cot A} = \dfrac{\sin^2 A}{\sin A – \cos A} = -\dfrac{\sin^2 A}{\cos A – \sin A}\)
  योग \(= \dfrac{\cos^2 A – \sin^2 A}{\cos A – \sin A} = \cos A + \sin A\)
  मान \(= \cos A + \sin A – \sin A = \cos A\)

निम्नांकित व्यंजक का मान ज्ञात कीजिए: \(\csc^4 A(1 – \cos^4 A) – 2\cot^2 A – 1\)

• \(\sin^2 A\)
• \(\csc^2 A\)
• \(1\)
• \(0\)

Explanation:

• \(1 – \cos^4 A = (1 – \cos^2 A)(1 + \cos^2 A) = \sin^2 A(1 + \cos^2 A)\)
  \(\operatorname{cosec}^4 A\times\sin^2 A(1 + \cos^2 A) = \operatorname{cosec}^2 A(1 + \cos^2 A) = \operatorname{cosec}^2 A + \cot^2 A\)
  मान \(= \operatorname{cosec}^2 A + \cot^2 A – 2\cot^2 A – 1 = \operatorname{cosec}^2 A – \cot^2 A – 1 = 0\)

यदि \(6\tan A(\tan A + 1) = 5 – \tan A\) है, दिया गया है कि \(0 < A < \dfrac{\pi}{2}\) है, तो \((\sin A + \cos A)\) का मान ज्ञात करें।

• \(3\sqrt{5}\)
• \(\dfrac{5}{\sqrt{3}}\)
• \(5\sqrt{3}\)
• \(\dfrac{3}{\sqrt{5}}\)

Explanation:

• माना \(t = \tan A\); \(6t^2 + 6t = 5 – t \Rightarrow 6t^2 + 7t – 5 = 0\)
  \(t = \dfrac{-7 \pm 13}{12} = \dfrac{1}{2}\) (धनात्मक मान)
  \(\sin A = \dfrac{1}{\sqrt{5}},\ \cos A = \dfrac{2}{\sqrt{5}}\)
  \(\sin A + \cos A = \dfrac{3}{\sqrt{5}}\)

यदि \((2\cos A + 1)(2\cos A – 1) = 0,\ 0^\circ < A \le 90^\circ\), है, तो \(A\) का मान ज्ञात कीजिए।

• \(90^\circ\)
• \(45^\circ\)
• \(30^\circ\)
• \(60^\circ\)

Explanation:

• \(4\cos^2 A – 1 = 0 \Rightarrow \cos A = \pm\dfrac{1}{2}\)
  \(0^\circ \lt A \le 90^\circ\) में \(\cos A\) धनात्मक है, अतः \(\cos A = \dfrac{1}{2} \Rightarrow A = 60^\circ\)

\(\sin\alpha – \sin\beta = ?\)

• \(2\cos\dfrac{\alpha+\beta}{2}\sin\dfrac{\alpha-\beta}{2}\)
• \(2\sin\dfrac{\alpha+\beta}{2}\sin\dfrac{\alpha-\beta}{2}\)
• \(2\cos\dfrac{\alpha-\beta}{2}\sin\dfrac{\alpha+\beta}{2}\)
• \(2\cos\dfrac{\alpha+\beta}{2}\cos\dfrac{\alpha-\beta}{2}\)

Explanation:

• माना \(\alpha = u + v,\ \beta = u – v\), जहाँ \(u = \dfrac{\alpha+\beta}{2},\ v = \dfrac{\alpha-\beta}{2}\)
  \(\sin\alpha – \sin\beta = \sin(u+v) – \sin(u-v) = 2\cos u\sin v\)
  \(= 2\cos\dfrac{\alpha+\beta}{2}\sin\dfrac{\alpha-\beta}{2}\)

यदि \(\sin(A + B) = \cos(A + B)\), तो \(\tan A\) का मान कितना होगा?

• \(\dfrac{1-\tan B}{1+\tan B}\)
• \(\dfrac{1+\tan B}{1-\tan B}\)
• \(\dfrac{1+\sec B}{1-\sec B}\)
• \(\dfrac{1-\csc B}{1+\csc B}\)

Explanation:

• \(\sin(A+B) = \cos(A+B) \Rightarrow \tan(A+B) = 1\)
  \(\dfrac{\tan A + \tan B}{1 – \tan A\tan B} = 1 \Rightarrow \tan A + \tan B = 1 – \tan A\tan B\)
  \(\tan A(1 + \tan B) = 1 – \tan B \Rightarrow \tan A = \dfrac{1 – \tan B}{1 + \tan B}\)

\(\sin^4\theta + \cos^4\theta\) को \(\sin\theta\) के रूप में इस प्रकार लिखा जा सकता है-

• \(2\sin^4\theta + 2\sin^2\theta – 1\)
• \(2\sin^4\theta – 2\sin^2\theta\)
• \(2\sin^4\theta – 2\sin^2\theta – 1\)
• \(2\sin^4\theta – 2\sin^2\theta + 1\)

Explanation:

• \(\sin^4\theta + \cos^4\theta = (\sin^2\theta + \cos^2\theta)^2 – 2\sin^2\theta\cos^2\theta = 1 – 2\sin^2\theta(1 – \sin^2\theta)\)
  \(= 1 – 2\sin^2\theta + 2\sin^4\theta = 2\sin^4\theta – 2\sin^2\theta + 1\)

\(\dfrac{\cos x – \sqrt{3}\sin x}{2} = ?\)

• \(\cos\left(\dfrac{\pi}{3} – x\right)\)
• \(\sin\left(\dfrac{\pi}{3} – x\right)\)
• \(\cos\left(\dfrac{\pi}{3} + x\right)\)
• \(\sin\left(\dfrac{\pi}{3} + x\right)\)

Explanation:

• \(\dfrac{\cos x – \sqrt{3}\sin x}{2} = \dfrac{1}{2}\cos x – \dfrac{\sqrt{3}}{2}\sin x\)
  \(= \cos\dfrac{\pi}{3}\cos x – \sin\dfrac{\pi}{3}\sin x = \cos\left(\dfrac{\pi}{3} + x\right)\)

\(\cos x\csc x – \sin x\sec x = ?\)

• \(\cot\dfrac{2x}{2}\)
• \(\tan 2x\)
• \(\cot 2x\)
• \(2\cot 2x\)

Explanation:

• \(\cos x\operatorname{cosec}x – \sin x\sec x = \cot x – \tan x = \dfrac{\cos^2 x – \sin^2 x}{\sin x\cos x}\)
  \(= \dfrac{\cos 2x}{\frac{1}{2}\sin 2x} = 2\cot 2x\)

\(\sin 10^\circ – \dfrac{4}{3}\sin^3 10^\circ = ?\)

• \(\dfrac{1}{3\sqrt{3}}\)
• \(\dfrac{1}{6}\)
• \(\dfrac{1}{2\sqrt{3}}\)
• \(\dfrac{3}{6}\)

Explanation:

• \(\sin 3\theta = 3\sin\theta – 4\sin^3\theta\), अतः \(\sin\theta – \dfrac{4}{3}\sin^3\theta = \dfrac{1}{3}\sin 3\theta\)
  \(\theta = 10^\circ \Rightarrow \dfrac{1}{3}\sin 30^\circ = \dfrac{1}{6}\)

निम्नलिखित को त्रिकोणमितीय अनुपातों के संदर्भ में इसका मान ज्ञात करने के लिए हल करें। \((\sin A + \cos A)(1 – \sin A\cos A)\)

• \(\sin^3 A + \cos^3 A\)
• \(\sin^3 A – \cos^2 A\)
• \([\cos A – \sin A][\sin^2 A + \cos^2 A]\)
• \(\sin^3 A – \cos^3 A\)

Explanation:

• \(a^3 + b^3 = (a + b)(a^2 – ab + b^2)\) में \(a = \sin A,\ b = \cos A\)
  \(\sin^2 A + \cos^2 A = 1\), अतः \(a^2 – ab + b^2 = 1 – \sin A\cos A\)
  \((\sin A + \cos A)(1 – \sin A\cos A) = \sin^3 A + \cos^3 A\)