Mixed|MATHS
यदि \(\sin\theta = \dfrac{1}{2}\) है, तो \((3\cos\theta – 4\cos^3\theta)\) का मान है-
- \(0\)
- \(1\)
- \(2\)
- \(-1\)
Explanation:
- \(3\cos\theta – 4\cos^3\theta = -(4\cos^3\theta – 3\cos\theta) = -\cos 3\theta\)
\(\sin\theta = \dfrac{1}{2} \Rightarrow \theta = 30^\circ\)
\(-\cos(3 \times 30^\circ) = -\cos 90^\circ = 0\)
उत्तर: 0
निम्नलिखित व्यंजक का मान ज्ञात कीजिए। \(5(\sin^4\theta + \cos^4\theta) + 3(\sin^6\theta + \cos^6\theta) + 19\sin^2\theta\cos^2\theta\)
- \(8\)
- \(5\)
- \(6\)
- \(7\)
Explanation:
- माना \(s^2c^2 = \sin^2\theta\cos^2\theta\)
\(\sin^4\theta + \cos^4\theta = 1 – 2s^2c^2\)
\(\sin^6\theta + \cos^6\theta = 1 – 3s^2c^2\)
व्यंजक \(= 5(1 – 2s^2c^2) + 3(1 – 3s^2c^2) + 19s^2c^2\)
\(= 8 – 10s^2c^2 – 9s^2c^2 + 19s^2c^2 = 8\)
उत्तर: 8
यदि \(\sin Y = x\) है, तो \(\cos 2Y\) (जहाँ \(0 \le Y \le 90^\circ\)) का मान क्या होगा?
- \((\sqrt{2}-1)x\)
- \(\sqrt{2}x\)
- \(1-2x\)
- \(1-2x^2\)
Explanation:
- \(\cos 2Y = 1 – 2\sin^2 Y\)
\(\sin Y = x \Rightarrow \cos 2Y = 1 – 2x^2\)
उत्तर: \(1 – 2x^2\)
यदि \(A + B = C\) है, तो \(A\tan B\tan C\) का मान क्या होगा?
- \(\tan C + \tan A – \tan B\)
- \(\tan C + \tan A + \tan B\)
- \(\tan A – \tan B – \tan C\)
- \(\tan C – \tan A – \tan B\)
Explanation:
- प्रश्न में “A” की जगह \(\tan A\) होना चाहिए (अर्थात् \(\tan A\tan B\tan C\))।
\(C = A + B \Rightarrow \tan C = \tan(A + B) = \dfrac{\tan A + \tan B}{1 – \tan A\tan B}\)
\(\tan C(1 – \tan A\tan B) = \tan A + \tan B\)
\(\tan C – \tan A\tan B\tan C = \tan A + \tan B\)
\(\tan A\tan B\tan C = \tan C – \tan A – \tan B\)
उत्तर: \(\tan C – \tan A – \tan B\)
यदि \(x = 8(\sin\theta + \cos\theta)\) और \(y = 9(\sin\theta – \cos\theta)\) है, तो \(\dfrac{x^2}{8^2} + \dfrac{y^2}{9^2}\) का मान क्या होगा?
- \(4\)
- \(6\)
- \(8\)
- \(2\)
Explanation:
- \(\dfrac{x^2}{8^2} = (\sin\theta + \cos\theta)^2,\quad \dfrac{y^2}{9^2} = (\sin\theta – \cos\theta)^2\)
योग \(= 2(\sin^2\theta + \cos^2\theta) = 2\)
उत्तर: 2
\(\theta = 135^\circ;\ \gamma = 15^\circ\). \(\cos(\theta)\sin(\gamma)\) का मान क्या होगा?
- \(\dfrac{\sqrt{3}-1}{2}\)
- \(\dfrac{1-\sqrt{3}}{2}\)
- \(2-\sqrt{3}\)
- \(\sqrt{3}-2\)
Explanation:
- \(\cos 135^\circ = -\dfrac{1}{\sqrt{2}}\), \(\sin 15^\circ = \dfrac{\sqrt{3} – 1}{2\sqrt{2}}\)
गुणनफल \(= -\dfrac{1}{\sqrt{2}} \times \dfrac{\sqrt{3} – 1}{2\sqrt{2}} = -\dfrac{\sqrt{3} – 1}{4} = \dfrac{1 – \sqrt{3}}{4}\)
उत्तर: \(\dfrac{1 – \sqrt{3}}{2}\) (उत्तर-कुंजी अनुसार)
नोट: सही मान \(\dfrac{1 – \sqrt{3}}{4} \approx -0.183\) आता है। कुंजी वाला विकल्प \(\dfrac{1 – \sqrt{3}}{2} \approx -0.366\) है, जिसमें हर 4 की जगह 2 छपा हुआ है।
\(\sin 75^\circ + \sin 15^\circ\) का मान क्या है?
- \(\dfrac{1}{\sqrt{2}}\)
- \(\dfrac{\sqrt{3}}{2}\)
- \(\sqrt{\dfrac{3}{2}}\)
- \(\dfrac{3}{\sqrt{2}}\)
Explanation:
- \(\sin 75^\circ + \sin 15^\circ = 2\sin\dfrac{75^\circ + 15^\circ}{2}\cos\dfrac{75^\circ – 15^\circ}{2}\)
\(= 2\sin 45^\circ\cos 30^\circ = 2 \times \dfrac{1}{\sqrt{2}} \times \dfrac{\sqrt{3}}{2} = \sqrt{\dfrac{3}{2}}\)
उत्तर: \(\sqrt{\dfrac{3}{2}}\)
\(\tan 27^\circ\tan 34^\circ + \tan 34^\circ\tan 29^\circ + \tan 29^\circ\tan 27^\circ = ?\)
- \(0\)
- \(-1\)
- \(\sqrt{3}\)
- \(1\)
Explanation:
- कोण \(27^\circ + 34^\circ + 29^\circ = 90^\circ\)
यदि \(A + B + C = 90^\circ\) हो तो \(\tan A\tan B + \tan B\tan C + \tan C\tan A = 1\)
(क्योंकि \(\tan(A + B) = \cot C\) से \(\tan A + \tan B = \cot C(1 – \tan A\tan B)\))
उत्तर: 1
यदि \(\tan 3\theta\cdot\tan 7\theta = 1\) है, जहां \(\theta\) एक न्यून कोण है, तो \(\cot 15\theta\) का मान ज्ञात कीजिए।
- \(1\)
- \(-1\)
- \(-\sqrt{3}\)
- \(\sqrt{3}\)
Explanation:
- \(\tan 3\theta \cdot \tan 7\theta = 1 \Rightarrow 3\theta + 7\theta = 90^\circ\)
\(\theta = 9^\circ\)
\(\cot 15\theta = \cot 135^\circ = \cot(180^\circ – 45^\circ) = -\cot 45^\circ = -1\)
उत्तर: \(-1\)
\(\csc 15^\circ\sec 15^\circ = ?\)
- \(0.5\)
- \(4\)
- \(2\)
- \(1\)
Explanation:
- \(\csc 15^\circ\sec 15^\circ = \dfrac{1}{\sin 15^\circ\cos 15^\circ} = \dfrac{2}{2\sin 15^\circ\cos 15^\circ} = \dfrac{2}{\sin 30^\circ}\)
\(= \dfrac{2}{1/2} = 4\)
उत्तर: 4
यदि \(\tan 40^\circ = \alpha\) है, तो \(\dfrac{\tan 320^\circ – \tan 310^\circ}{1 + \tan 320^\circ\cdot\tan 310^\circ}\) का मान ज्ञात कीजिए।
- \(\dfrac{1-\alpha^2}{\alpha}\)
- \(\dfrac{1+\alpha^2}{2\alpha}\)
- \(\dfrac{1-\alpha^2}{2\alpha}\)
- \(\dfrac{1+\alpha^2}{\alpha}\)
Explanation:
- \(\tan 320^\circ = \tan(360^\circ – 40^\circ) = -\tan 40^\circ = -\alpha\)
\(\tan 310^\circ = \tan(360^\circ – 50^\circ) = -\tan 50^\circ = -\cot 40^\circ = -\dfrac{1}{\alpha}\)
व्यंजक \(= \dfrac{-\alpha + \frac{1}{\alpha}}{1 + \alpha \cdot \frac{1}{\alpha}} = \dfrac{\frac{1 – \alpha^2}{\alpha}}{2} = \dfrac{1 – \alpha^2}{2\alpha}\)
उत्तर: \(\dfrac{1 + \alpha^2}{2\alpha}\) (उत्तर-कुंजी अनुसार)
नोट: गणना से सही मान \(\dfrac{1 – \alpha^2}{2\alpha}\) आता है (जो विकल्प 3 है), जबकि कुंजी में विकल्प 2 \(\dfrac{1 + \alpha^2}{2\alpha}\) अंकित है। संख्यात्मक जाँच: \(\alpha = 0.8391\) पर व्यंजक \(= 0.1763\), विकल्प 3 भी \(0.1763\), पर विकल्प 2 \(= 1.0154\)।
यदि \(\tan(\alpha+\beta) = a,\ \tan(\alpha-\beta) = b\) है, तो \(\tan 2\alpha\) का मान ज्ञात कीजिए।
- \(\dfrac{a+b}{1-ab}\)
- \(\dfrac{a+b}{1+ab}\)
- \(\dfrac{a-b}{1+ab}\)
- \(\dfrac{a-b}{1-ab}\)
Explanation:
- \(2\alpha = (\alpha + \beta) + (\alpha – \beta)\)
\(\tan 2\alpha = \dfrac{\tan(\alpha + \beta) + \tan(\alpha – \beta)}{1 – \tan(\alpha + \beta)\tan(\alpha – \beta)} = \dfrac{a + b}{1 – ab}\)
उत्तर: \(\dfrac{a + b}{1 – ab}\)
यदि \(A\) एक न्यून कोण है, तो निम्न का सरलीकृत रूप क्या होगा? \(\dfrac{\cos(\pi – A)\cdot\cot\left(\frac{\pi}{2} + A\right)\cos(-A)}{\tan(\pi + A)\tan\left(\frac{3\pi}{2} + A\right)\sin(2\pi – A)}\)
- \(\cos^2 A\)
- \(\sin A\)
- \(\sin^2 A\)
- \(\cos A\)
Explanation:
- \(\cos(\pi – A) = -\cos A,\quad \cot\left(\dfrac{\pi}{2} + A\right) = -\tan A,\quad \cos(-A) = \cos A\)
\(\tan(\pi + A) = \tan A,\quad \tan\left(\dfrac{3\pi}{2} + A\right) = -\cot A,\quad \sin(2\pi – A) = -\sin A\)
अंश \(= (-\cos A)(-\tan A)(\cos A) = \sin A\cos A\)
हर \(= (\tan A)(-\cot A)(-\sin A) = \sin A\)
मान \(= \dfrac{\sin A\cos A}{\sin A} = \cos A\)
उत्तर: \(\cos A\)
यदि \(a\cot\theta + b\csc\theta = p\) और \(b\cot\theta + a\csc\theta = q\) है, तो \(p^2 – q^2\) निम्न में से किसके बराबर है?
- \(a^2 + b^2\)
- \(a^2 – b^2\)
- \(b^2 – a^2\)
- \(-a\)
Explanation:
- \(p^2 – q^2 = (a\cot\theta + b\csc\theta)^2 – (b\cot\theta + a\csc\theta)^2\)
\(= a^2\cot^2\theta + b^2\csc^2\theta – b^2\cot^2\theta – a^2\csc^2\theta\)
\(= (a^2 – b^2)(\cot^2\theta – \csc^2\theta) = (a^2 – b^2)(-1) = b^2 – a^2\)
उत्तर: \(b^2 – a^2\)
\(\cos(A+B) = \cos A\cos B – \sin A\sin B\) का उपयोग करके, \(\cos 75^\circ\) का मान ज्ञात कीजिए।
- \(\dfrac{\sqrt{5}-1}{4}\)
- \(\dfrac{\sqrt{5}+1}{4}\)
- \(\dfrac{\sqrt{6}-\sqrt{2}}{4}\)
- \(\dfrac{\sqrt{6}+\sqrt{2}}{4}\)
Explanation:
- \(\cos 75^\circ = \cos(45^\circ + 30^\circ) = \cos 45^\circ\cos 30^\circ – \sin 45^\circ\sin 30^\circ\)
\(= \dfrac{1}{\sqrt{2}} \times \dfrac{\sqrt{3}}{2} – \dfrac{1}{\sqrt{2}} \times \dfrac{1}{2} = \dfrac{\sqrt{3} – 1}{2\sqrt{2}} = \dfrac{\sqrt{6} – \sqrt{2}}{4}\)
उत्तर: \(\dfrac{\sqrt{6} – \sqrt{2}}{4}\)
त्रिकोणमितीय सूत्रों का उपयोग करके \(\left(\dfrac{\sin(x-y)}{\sin(x+y)}\right)\left(\dfrac{\tan x + \tan y}{\tan x – \tan y}\right)\) का मान ज्ञात करें।
- \(-2\)
- \(2\)
- \(0\)
- \(1\)
Explanation:
- \(\tan x + \tan y = \dfrac{\sin(x + y)}{\cos x\cos y},\quad \tan x – \tan y = \dfrac{\sin(x – y)}{\cos x\cos y}\)
अतः \(\dfrac{\tan x + \tan y}{\tan x – \tan y} = \dfrac{\sin(x + y)}{\sin(x – y)}\)
गुणनफल \(= \dfrac{\sin(x – y)}{\sin(x + y)} \times \dfrac{\sin(x + y)}{\sin(x – y)} = 1\)
उत्तर: 1
यदि \(\tan\dfrac{A}{2} = x\) है, तो \(x\) का मान ज्ञात कीजिए।
- \(\sqrt{\dfrac{1+\cos A}{1-\cos A}}\)
- \(\sqrt{\dfrac{1-\sin A}{1+\sin A}}\)
- \(\sqrt{\dfrac{1-\cos A}{1+\cos A}}\)
- \(\sqrt{\dfrac{\cos A-1}{1+\cos A}}\)
Explanation:
- \(\tan^2\dfrac{A}{2} = \dfrac{1 – \cos A}{1 + \cos A}\) (अर्ध-कोण सूत्र)
\(x = \tan\dfrac{A}{2} = \sqrt{\dfrac{1 – \cos A}{1 + \cos A}}\)
उत्तर: \(\sqrt{\dfrac{1 – \cos A}{1 + \cos A}}\)
निम्नलिखित का मान ज्ञात कीजिए- \(\sin 2x + 2\sin 4x + \sin 6x\)
- \(4\cos^2 x\sin 4x\)
- \(4\cos^2 x\sin x\)
- \(2\cos^2 x\sin 4x\)
- \(4\sin^2 x\sin 4x\)
Explanation:
- \(\sin 2x + \sin 6x = 2\sin 4x\cos 2x\)
व्यंजक \(= 2\sin 4x\cos 2x + 2\sin 4x = 2\sin 4x(1 + \cos 2x)\)
\(1 + \cos 2x = 2\cos^2 x\)
\(= 4\cos^2 x\sin 4x\)
उत्तर: \(4\cos^2 x\sin 4x\)
\(\tan(A-B) = \dfrac{\tan A – \tan B}{1 + \tan A\tan B}\) के उपयोग से, \(\tan 15^\circ\) का मान ज्ञात कीजिए।
- \(\sqrt{3}+1\)
- \(\sqrt{3}-1\)
- \(2-\sqrt{3}\)
- \(2+\sqrt{3}\)
Explanation:
- \(\tan 15^\circ = \tan(45^\circ – 30^\circ) = \dfrac{\tan 45^\circ – \tan 30^\circ}{1 + \tan 45^\circ\tan 30^\circ}\)
\(= \dfrac{1 – \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}} = \dfrac{\sqrt{3} – 1}{\sqrt{3} + 1}\)
\(= \dfrac{(\sqrt{3} – 1)^2}{3 – 1} = \dfrac{4 – 2\sqrt{3}}{2} = 2 – \sqrt{3}\)
उत्तर: \(2 – \sqrt{3}\)
\(\cot(A-B) = \dfrac{\cot A\cot B + 1}{\cot B – \cot A}\) के उपयोग से, \(\cot 15^\circ\) का मान ज्ञात कीजिए।
- \(2+\sqrt{3}\)
- \(2-\sqrt{3}\)
- \(\sqrt{3}-1\)
- \(\sqrt{3}+1\)
Explanation:
- \(\cot 15^\circ = \cot(45^\circ – 30^\circ) = \dfrac{\cot 45^\circ\cot 30^\circ + 1}{\cot 30^\circ – \cot 45^\circ}\)
\(= \dfrac{\sqrt{3} + 1}{\sqrt{3} – 1} = \dfrac{(\sqrt{3} + 1)^2}{3 – 1} = \dfrac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3}\)
उत्तर: \(2 + \sqrt{3}\)
\(\sqrt{\dfrac{1-\sin 3\theta}{1+\sin 3\theta}} = ?\)
- \(\sec 3\theta – \tan 3\theta\)
- \((\sec 3\theta – \tan 3\theta)^3\)
- \((\sec 3\theta – \tan 3\theta)^2\)
- \(\sec 3\theta + \tan 3\theta\)
Explanation:
- \(\sqrt{\dfrac{1 – \sin 3\theta}{1 + \sin 3\theta}} = \sqrt{\dfrac{(1 – \sin 3\theta)^2}{1 – \sin^2 3\theta}} = \dfrac{1 – \sin 3\theta}{\cos 3\theta}\)
\(= \sec 3\theta – \tan 3\theta\)
उत्तर: \(\sec 3\theta – \tan 3\theta\)
यदि \(\dfrac{\cos\alpha}{\sin\beta} = 10\) और \(\dfrac{\cos\alpha}{\cos\beta} = 11\) है, तो \(\cos^2\beta\) का मान क्या होगा?
- \(\dfrac{121}{132}\)
- \(\dfrac{100}{221}\)
- \(\dfrac{88}{108}\)
- \(\dfrac{221}{121}\)
Explanation:
- \(\sin\beta = \dfrac{\cos\alpha}{10},\quad \cos\beta = \dfrac{\cos\alpha}{11}\)
\(\sin^2\beta + \cos^2\beta = 1 \Rightarrow \cos^2\alpha\left(\dfrac{1}{100} + \dfrac{1}{121}\right) = 1\)
\(\cos^2\alpha \times \dfrac{221}{12100} = 1 \Rightarrow \cos^2\alpha = \dfrac{12100}{221}\)
\(\cos^2\beta = \dfrac{\cos^2\alpha}{121} = \dfrac{100}{221}\)
उत्तर: \(\dfrac{100}{221}\)
Simplify: \(\dfrac{1+\sin t}{4-4\sin t} – \dfrac{1-\sin t}{4+4\sin t}\)
- \(4\tan t\cdot\sin t\)
- \(\tan t\cdot\sec t\)
- \(\tan t – \sin t\)
- \(\tan t \div \sin t\)
Explanation:
- \(\dfrac{1 + \sin t}{4(1 – \sin t)} – \dfrac{1 – \sin t}{4(1 + \sin t)} = \dfrac{(1 + \sin t)^2 – (1 – \sin t)^2}{4(1 – \sin^2 t)}\)
\(= \dfrac{4\sin t}{4\cos^2 t} = \dfrac{\sin t}{\cos t} \cdot \dfrac{1}{\cos t} = \tan t\sec t\)
उत्तर: \(\tan t \cdot \sec t\)
यदि \(a\cot\theta = b\) है, तो \(\dfrac{b\cos\theta + a\sin\theta}{b\cos\theta – a\sin\theta}\) का मान क्या होगा?
- \(\dfrac{b^2+a^2}{b^2-a^2}\)
- \(b^2 + a^2\)
- \(\dfrac{b^2-a^2}{b^2+a^2}\)
- \(0\)
Explanation:
- \(a\cot\theta = b \Rightarrow \cot\theta = \dfrac{b}{a}\)
अंश और हर को \(\sin\theta\) से भाग देने पर \(\dfrac{b\cot\theta + a}{b\cot\theta – a} = \dfrac{\frac{b^2}{a} + a}{\frac{b^2}{a} – a}\)
\(= \dfrac{b^2 + a^2}{b^2 – a^2}\)
उत्तर: \(\dfrac{b^2 + a^2}{b^2 – a^2}\)
\(\csc(\alpha+\beta) = \dfrac{\sec\alpha\times\sec\beta\times\csc\alpha\times\csc\beta}{\sec\alpha\times\csc\beta + \csc\alpha\times\sec\beta}\) के उपयोग से, \(\csc 75^\circ\) का मान ज्ञात कीजिए।
- \(\dfrac{\sqrt{6}+\sqrt{2}}{4}\)
- \(\dfrac{\sqrt{6}-\sqrt{2}}{4}\)
- \(\sqrt{6}-\sqrt{2}\)
- \(\sqrt{6}+\sqrt{2}\)
Explanation:
- \(\alpha = 45^\circ,\ \beta = 30^\circ\); \(\sec 45^\circ = \csc 45^\circ = \sqrt{2}\), \(\sec 30^\circ = \dfrac{2}{\sqrt{3}}\), \(\csc 30^\circ = 2\)
अंश \(= \sqrt{2} \times \dfrac{2}{\sqrt{3}} \times \sqrt{2} \times 2 = \dfrac{8}{\sqrt{3}}\)
हर \(= \sec\alpha\csc\beta + \csc\alpha\sec\beta = 2\sqrt{2} + \dfrac{2\sqrt{2}}{\sqrt{3}} = \dfrac{2\sqrt{2}(\sqrt{3} + 1)}{\sqrt{3}}\)
\(\csc 75^\circ = \dfrac{8}{2\sqrt{2}(\sqrt{3} + 1)} = \dfrac{2\sqrt{2}}{\sqrt{3} + 1} = \dfrac{2\sqrt{2}(\sqrt{3} – 1)}{2} = \sqrt{6} – \sqrt{2}\)
उत्तर: \(\sqrt{6} – \sqrt{2}\)
यदि \(\dfrac{21\cos A + 3\sin A}{3\cos A + 4\sin A} = 2\) है, तो \(\cot A\) का मान ज्ञात कीजिए।
- \(\dfrac{9}{11}\)
- \(\dfrac{11}{9}\)
- \(\dfrac{1}{3}\)
- \(\dfrac{11}{10}\)
Explanation:
- \(21\cos A + 3\sin A = 2(3\cos A + 4\sin A) = 6\cos A + 8\sin A\)
\(15\cos A = 5\sin A \Rightarrow \tan A = 3\)
\(\cot A = \dfrac{1}{3}\)
उत्तर: \(\dfrac{1}{3}\)
निम्नलिखित का मान ज्ञात कीजिये: \(\sqrt{2+\sqrt{2+\sqrt{2+2\cos 8\theta}}}\)
- \(2\cos\theta\)
- \(2\cos 2\theta\)
- \(\sin 2\theta\)
- \(\cos 2\theta\)
Explanation:
- \(2 + 2\cos 8\theta = 4\cos^2 4\theta \Rightarrow \sqrt{\ } = 2\cos 4\theta\)
\(2 + 2\cos 4\theta = 4\cos^2 2\theta \Rightarrow \sqrt{\ } = 2\cos 2\theta\)
\(2 + 2\cos 2\theta = 4\cos^2\theta \Rightarrow \sqrt{\ } = 2\cos\theta\)
उत्तर: \(2\cos\theta\)
\(\dfrac{2+\tan^2 2A+\cot^2 2A}{\sec 2A\csc 2A} = ?\)
- \(\sec 2A + \tan 2A\)
- \(\sec 2A + \csc 2A\)
- \(\cot 2A – \csc 2A\)
- \(\sec 2A\csc 2A\)
Explanation:
- \(2 + \tan^2 2A + \cot^2 2A = (\tan 2A + \cot 2A)^2\)
\(\tan 2A + \cot 2A = \dfrac{\sin^2 2A + \cos^2 2A}{\sin 2A\cos 2A} = \sec 2A\csc 2A\)
व्यंजक \(= \dfrac{(\sec 2A\csc 2A)^2}{\sec 2A\csc 2A} = \sec 2A\csc 2A\)
उत्तर: \(\sec 2A\csc 2A\)
व्यंजक \(\dfrac{1-\sin(2t)}{1+\sin(2t)}\times\dfrac{\cos(t)+\sin(t)}{\cos(t)-\sin(t)}\) का मान है:
- \(\dfrac{1-2\tan(t)}{1+2\tan(t)}\)
- \(\dfrac{1-\tan(t)}{1+\tan(t)}\)
- \(\dfrac{1+2\tan(t)}{1-2\tan(t)}\)
- \(\dfrac{1+\tan(t)}{1-\tan(t)}\)
Explanation:
- \(1 – \sin 2t = (\cos t – \sin t)^2,\quad 1 + \sin 2t = (\cos t + \sin t)^2\)
व्यंजक \(= \dfrac{(\cos t – \sin t)^2}{(\cos t + \sin t)^2} \times \dfrac{\cos t + \sin t}{\cos t – \sin t} = \dfrac{\cos t – \sin t}{\cos t + \sin t}\)
\(\cos t\) से भाग देने पर \(= \dfrac{1 – \tan t}{1 + \tan t}\)
उत्तर: \(\dfrac{1 – \tan t}{1 + \tan t}\)
यदि \(m = a\sec A\) और \(y = b\tan A\) है, तो \(b^2m^2 – a^2y^2 + \dfrac{a^2y^2}{b^2m^2} + \cos^2 A\) का मान ज्ञात कीजिए।
- \(a^2b^2\)
- \(1-a^2b^2\)
- \(a^2b^2+2\)
- \(a^2b^2+1\)
Explanation:
- \(b^2m^2 – a^2y^2 = b^2a^2\sec^2 A – a^2b^2\tan^2 A = a^2b^2(\sec^2 A – \tan^2 A) = a^2b^2\)
\(\dfrac{a^2y^2}{b^2m^2} = \dfrac{a^2b^2\tan^2 A}{b^2a^2\sec^2 A} = \sin^2 A\)
कुल \(= a^2b^2 + \sin^2 A + \cos^2 A = a^2b^2 + 1\)
उत्तर: \(a^2b^2 + 1\)
यदि \(a\sin^3 x + b\cos^3 x = \sin x\cos x\) और \(a\sin x = b\cos x\) हो, तो \(a^2 + b^2\) का मान ज्ञात कीजिए, बशर्ते कि \(x\) न तो \(0^\circ\) और न ही \(90^\circ\) है।
- \(0\)
- \(1\)
- \(a^2 – b^2\)
- \(a^2 + b^2\)
Explanation:
- \(a\sin x = b\cos x\) का उपयोग: \(a\sin^3 x + b\cos^3 x = a\sin x \cdot \sin^2 x + b\cos x \cdot \cos^2 x\)
\(= a\sin x(\sin^2 x + \cos^2 x) = a\sin x\)
\(a\sin x = \sin x\cos x \Rightarrow a = \cos x\)
\(b\cos x = a\sin x = \sin x\cos x \Rightarrow b = \sin x\)
\(a^2 + b^2 = \cos^2 x + \sin^2 x = 1\)
उत्तर: 1
यदि \(\sin A + \cos A = \dfrac{1}{2\sqrt{2}}\), तो \(\sin^4 A + \cos^4 A\) का मान ज्ञात कीजिए।
- \(79/128\)
- \(7/13\)
- \(79/126\)
- \(5/13\)
Explanation:
- वर्ग करने पर \(1 + 2\sin A\cos A = \dfrac{1}{8} \Rightarrow \sin A\cos A = -\dfrac{7}{16}\)
\(\sin^4 A + \cos^4 A = 1 – 2\sin^2 A\cos^2 A = 1 – 2 \times \dfrac{49}{256}\)
\(= 1 – \dfrac{49}{128} = \dfrac{79}{128}\)
उत्तर: \(\dfrac{79}{128}\)
माना कि \(x = r\cos(t),\ y = r\sin(t)\cos(u),\ z = r\sin(t)\sin(u)\). तो \(x^2 + y^2 + z^2\) का मान ज्ञात करें।
- \(r^2\cos(u)\)
- \(r^2\)
- \(2r\)
- \(r^2\sin(t)\)
Explanation:
- \(x^2 + y^2 + z^2 = r^2\cos^2 t + r^2\sin^2 t\cos^2 u + r^2\sin^2 t\sin^2 u\)
\(= r^2\cos^2 t + r^2\sin^2 t(\cos^2 u + \sin^2 u) = r^2(\cos^2 t + \sin^2 t) = r^2\)
उत्तर: \(r^2\)
\((\sin A + \csc A)^2 + (\cos A + \sec A)^2\) का मान ज्ञात कीजिए:
- \(3 + \tan^2 A + \cot^2 A\)
- \(7 + \tan^2 A + \cot^2 A\)
- \(5 + \tan^2 A + \cot^2 A\)
- \(1 + \tan^2 A + \cot^2 A\)
Explanation:
- \((\sin A + \csc A)^2 = \sin^2 A + 2 + \csc^2 A\)
\((\cos A + \sec A)^2 = \cos^2 A + 2 + \sec^2 A\)
योग \(= 1 + 4 + \csc^2 A + \sec^2 A = 5 + (1 + \cot^2 A) + (1 + \tan^2 A)\)
\(= 7 + \tan^2 A + \cot^2 A\)
उत्तर: \(7 + \tan^2 A + \cot^2 A\)
\(\dfrac{\sin(A+B)}{\sin A\cos B} = ?\)
- \(1 + \cot A\tan B\)
- \(1 + \tan A\cot B\)
- \(1 – \sin A\cos B\)
- \(1 – \cot A\tan B\)
Explanation:
- \(\dfrac{\sin(A + B)}{\sin A\cos B} = \dfrac{\sin A\cos B + \cos A\sin B}{\sin A\cos B}\)
\(= 1 + \dfrac{\cos A}{\sin A} \cdot \dfrac{\sin B}{\cos B} = 1 + \cot A\tan B\)
उत्तर: \(1 + \cot A\tan B\)
निम्न का मान ज्ञात कीजिए। \(\dfrac{\sin 8\theta\cos\theta – \sin 6\theta\cos 3\theta}{\cos 2\theta\cos\theta – \sin 3\theta\sin 4\theta}\)
- \(\cot\theta\)
- \(\tan 2\theta\)
- \(\tan\theta\)
- \(\cot 2\theta\)
Explanation:
- अंश: \(\sin 8\theta\cos\theta = \dfrac{1}{2}(\sin 9\theta + \sin 7\theta)\), \(\sin 6\theta\cos 3\theta = \dfrac{1}{2}(\sin 9\theta + \sin 3\theta)\)
अंश \(= \dfrac{1}{2}(\sin 7\theta – \sin 3\theta) = \cos 5\theta\sin 2\theta\)
हर: \(\cos 2\theta\cos\theta = \dfrac{1}{2}(\cos 3\theta + \cos\theta)\), \(\sin 3\theta\sin 4\theta = \dfrac{1}{2}(\cos\theta – \cos 7\theta)\)
हर \(= \dfrac{1}{2}(\cos 3\theta + \cos 7\theta) = \cos 5\theta\cos 2\theta\)
मान \(= \tan 2\theta\)
उत्तर: \(\tan 2\theta\)
यदि \(\left\{\left(\dfrac{\sec\theta-1}{\sec\theta+1}\right)\right\}^n = \csc\theta – \cot\theta\), तो \(n = ?\)
- \(1\)
- \(0.5\)
- \(-1\)
- \(-0.5\)
Explanation:
- \(\dfrac{\sec\theta – 1}{\sec\theta + 1} = \dfrac{1 – \cos\theta}{1 + \cos\theta} = \tan^2\dfrac{\theta}{2}\)
\(\csc\theta – \cot\theta = \dfrac{1 – \cos\theta}{\sin\theta} = \tan\dfrac{\theta}{2}\)
\(\left(\tan^2\dfrac{\theta}{2}\right)^n = \tan\dfrac{\theta}{2} \Rightarrow 2n = 1 \Rightarrow n = 0.5\)
उत्तर: 0.5
\((2\cos^2\theta – 1)\left[\dfrac{1+\tan\theta}{1-\tan\theta} + \dfrac{1-\tan\theta}{1+\tan\theta}\right]\) का मान क्या है?
- \(2\)
- \(0\)
- \(\dfrac{\sqrt{3}}{2}\)
- \(1\)
Explanation:
- \(2\cos^2\theta – 1 = \cos 2\theta\)
कोष्ठक \(= \dfrac{(1 + \tan\theta)^2 + (1 – \tan\theta)^2}{1 – \tan^2\theta} = \dfrac{2(1 + \tan^2\theta)}{1 – \tan^2\theta} = \dfrac{2}{\cos 2\theta}\)
(क्योंकि \(\dfrac{1 – \tan^2\theta}{1 + \tan^2\theta} = \cos 2\theta\))
गुणनफल \(= \cos 2\theta \times \dfrac{2}{\cos 2\theta} = 2\)
उत्तर: 2
यदि \(\cos A + \cos B + \cos C = 3\), then what is the value of \(\sin A + \sin B + \sin C\) का मान क्या है?
- \(1\)
- \(2\)
- \(0\)
- \(-1\)
Explanation:
- प्रत्येक \(\cos\) का अधिकतम मान 1 है, इसलिए तीनों का योग 3 तभी होगा जब \(\cos A = \cos B = \cos C = 1\)
अतः \(A = B = C = 0^\circ\) और \(\sin A = \sin B = \sin C = 0\)
योग \(= 0\)
उत्तर: 0
यदि \(\cos^4 a – \sin^4 a = \dfrac{5}{6}\), तो \(2\cos^2 a – 1\) का मान ज्ञात करें।
- \(6/11\)
- \(5/6\)
- \(6/5\)
- \(11/6\)
Explanation:
- \(\cos^4 a – \sin^4 a = (\cos^2 a – \sin^2 a)(\cos^2 a + \sin^2 a) = \cos^2 a – \sin^2 a\)
\(\cos^2 a – \sin^2 a = 2\cos^2 a – 1 = \dfrac{5}{6}\)
उत्तर: \(\dfrac{5}{6}\)
यदि \(\cos A = \sin^2 A\), और \(a\sin^{12}A + b\sin^{10}A + c\sin^8 A + \sin^6 A = 1\), तो \(a + b + c = ?\)
- \(7\)
- \(8\)
- \(9\)
- \(6\)
Explanation:
- \(\cos A = \sin^2 A\); दोनों पक्षों का वर्ग करने पर \(\cos^2 A = \sin^4 A\)
\(1 – \sin^2 A = \sin^4 A \Rightarrow \sin^4 A + \sin^2 A = 1\)
\(\sin^2 A(\sin^2 A + 1) = 1 \Rightarrow \sin^2 A + 1 = \dfrac{1}{\sin^2 A}\)
दोनों पक्षों का घन करने पर \(\sin^6 A(1 + \sin^2 A)^3 = 1\)
\(\sin^6 A + 3\sin^8 A + 3\sin^{10} A + \sin^{12} A = 1\)
दिए गए समीकरण से तुलना करने पर \(a = 1,\ b = 3,\ c = 3\), अतः \(a + b + c = 7\)
उत्तर: 7
\(\sqrt{\dfrac{1-\tan A}{1+\tan A}} = ?\)
- \(\sqrt{\dfrac{1+\sin 2A}{\cos 2A}}\)
- \(\sqrt{\dfrac{1-\sin 2A}{\cos 2A}}\)
- \(\sqrt{\dfrac{1+\sin A}{\cos A}}\)
- \(\sqrt{\dfrac{1-\sin A}{\cos A}}\)
Explanation:
- \(\sqrt{\dfrac{1 – \tan A}{1 + \tan A}} = \sqrt{\dfrac{\cos A – \sin A}{\cos A + \sin A}}\)
अंश और हर को \((\cos A – \sin A)\) से गुणा करने पर \(\sqrt{\dfrac{(\cos A – \sin A)^2}{\cos^2 A – \sin^2 A}} = \sqrt{\dfrac{1 – \sin 2A}{\cos 2A}}\)
उत्तर: \(\sqrt{\dfrac{1 – \sin 2A}{\cos 2A}}\)
\((1 + \cot A – \csc A)(1 + \tan A + \sec A) – 3(\sin^2 A + \cos^2 A)\) का मान ज्ञात कीजिए।
- \(-1\)
- \(1\)
- \(2\)
- \(-2\)
Explanation:
- \(1 + \cot A – \csc A = \dfrac{\sin A + \cos A – 1}{\sin A}\)
\(1 + \tan A + \sec A = \dfrac{\cos A + \sin A + 1}{\cos A}\)
गुणनफल \(= \dfrac{(\sin A + \cos A)^2 – 1}{\sin A\cos A} = \dfrac{2\sin A\cos A}{\sin A\cos A} = 2\)
मान \(= 2 – 3(\sin^2 A + \cos^2 A) = 2 – 3 = -1\)
उत्तर: \(-1\)
निम्नलिखित व्यंजक का मान ज्ञात कीजिए। \((1 + \tan\theta + \sec\theta)\times(1 + \cot\theta – \csc\theta)\)
- \(1\)
- \(\cot\theta\csc\theta\)
- \(2\)
- \(\tan\theta\)
Explanation:
- \(1 + \tan\theta + \sec\theta = \dfrac{\cos\theta + \sin\theta + 1}{\cos\theta}\)
\(1 + \cot\theta – \csc\theta = \dfrac{\sin\theta + \cos\theta – 1}{\sin\theta}\)
गुणनफल \(= \dfrac{(\sin\theta + \cos\theta)^2 – 1}{\sin\theta\cos\theta} = \dfrac{2\sin\theta\cos\theta}{\sin\theta\cos\theta} = 2\)
उत्तर: 2
यदि \(A, B\) और \(C,\ \triangle ABC\) के आंतरिक कोण हैं, तो \(\left\{\tan\dfrac{A}{2} + \csc\dfrac{B+C}{2}\right\}\left\{\tan\dfrac{A}{2} – \csc\dfrac{B+C}{2}\right\}\) का मान क्या है?
- \(2\)
- \(-1\)
- \(-2\)
- \(1\)
Explanation:
- \(A + B + C = 180^\circ \Rightarrow \dfrac{B + C}{2} = 90^\circ – \dfrac{A}{2}\)
\(\csc\dfrac{B + C}{2} = \csc\left(90^\circ – \dfrac{A}{2}\right) = \sec\dfrac{A}{2}\)
व्यंजक \(= \tan^2\dfrac{A}{2} – \sec^2\dfrac{A}{2} = -1\)
उत्तर: \(-1\)
\(\dfrac{(\sec\theta+\tan\theta)(1-\sin\theta)}{\csc\theta(1+\cos\theta)(\csc\theta-\cot\theta)} = ?\)
- \(\sin\theta\)
- \(\sec\theta\)
- \(\cos\theta\)
- \(\csc\theta\)
Explanation:
- \((\sec\theta + \tan\theta)(1 – \sin\theta) = \dfrac{(1 + \sin\theta)(1 – \sin\theta)}{\cos\theta} = \dfrac{\cos^2\theta}{\cos\theta} = \cos\theta\)
\(\csc\theta – \cot\theta = \dfrac{1 – \cos\theta}{\sin\theta}\)
हर \(= \csc\theta(1 + \cos\theta) \cdot \dfrac{1 – \cos\theta}{\sin\theta} = \dfrac{1}{\sin\theta} \cdot \dfrac{\sin^2\theta}{\sin\theta} = 1\)
मान \(= \cos\theta\)
उत्तर: \(\cos\theta\)
If \(\sqrt{\dfrac{1-\cos\theta}{1+\cos\theta}}\times\sqrt{\dfrac{\csc\theta-\cot\theta}{\csc\theta+\cot\theta}} = \dfrac{1-r}{1+r}\), then the value of r is:
- \(\sin\theta\)
- \(\csc\theta\)
- \(\sec\theta\)
- \(\cos\theta\)
Explanation:
\(0^\circ < \theta < 90^\circ\) के लिए, अगर \(\dfrac{\sec\theta(1-\sin\theta)(\sec\theta+\tan\theta)}{(\sec\theta-\tan\theta)^2} = \dfrac{1+k}{1-k}\) है, तो k बराबर है:
- \(\csc\theta\)
- \(\cos\theta\)
- \(\sec\theta\)
- \(\sin\theta\)
Explanation:
यदि \(\dfrac{\cos\alpha}{\sin\alpha+\cos\beta} + \dfrac{\cos\beta}{\sin\beta-\cos\alpha} = \dfrac{x}{\sin\alpha-\cos\beta} + \dfrac{\cos\beta}{\sin\beta+\cos\alpha}\) है, तो ‘x’ का मान है?
- \(\cos\beta\)
- \(\cos\alpha\)
- \(\sin\beta\)
- \(\sin\alpha\)
Explanation:
यह दिया गया है कि, \(\sqrt{\dfrac{1-\sin x}{1+\sin x}} = a – \tan x\) है तो a किसके समान है?
- \(\cos x\)
- \(\sin x\)
- \(\csc x\)
- \(\sec x\)
Explanation:
सभी \(\alpha_i\)’s के लिए, (\(i = 1, 2, 3, \ldots, 20\)) \(0^\circ\) और \(90^\circ\) के बीच स्थित है। यदि दिया गया है, तो \(\sin\alpha_1 + \sin\alpha_2 + \sin\alpha_3 + \ldots + \sin\alpha_{20} = 20\). \((\alpha_1 + \alpha_2 + \alpha_3 + \ldots + \alpha_{20})\) का (डिग्री में) मान क्या है?
- \(1800\)
- \(900\)
- \(0\)
- \(20\)
Explanation:
सभी \(\alpha_i\)’s के लिए, (\(i = 1, 2, 3, \ldots, 20\)) \(0^\circ\) और \(90^\circ\) के बीच स्थित है। यदि दिया गया है, तो \(\cos\alpha_1 + \cos\alpha_2 + \cos\alpha_3 + \ldots + \cos\alpha_{20} = 20\). \((\alpha_1 + \alpha_2 + \alpha_3 + \ldots + \alpha_{20})\) का (डिग्री में) मान क्या है?
- \(900^\circ\)
- \(1800^\circ\)
- \(0^\circ\)
- \(20^\circ\)
Explanation:
यदि \(A = 2(\sin^6\theta + \cos^6\theta) – 3(\sin^4\theta + \cos^4\theta)\) है तथा \(\cos\alpha = \sqrt{\dfrac{3+A}{5+A}}\) करने के लिए \(3\alpha\) का मान ज्ञात करें।
- \(45^\circ\)
- \(135^\circ\)
- \(180^\circ\)
- \(90^\circ\)
Explanation:
अगर \(\cos A,\ \sin A,\ \cot A\) गुणोत्तर श्रेणी में है, तो \(\tan^6 A – \tan^2 A\) का मान क्या होगा।
- \(\dfrac{1}{2}\)
- \(3\)
- \(\dfrac{1}{3}\)
- \(1\)
Explanation:
अगर \(\dfrac{1+\sin\theta-\cos\theta}{1+\sin\theta+\cos\theta} + \dfrac{1+\sin\theta+\cos\theta}{1+\sin\theta-\cos\theta} = 4\) है, तो निम्नलिखित में से कौन सा मान \(\theta\) के लिए, सही होगा?
- \(90^\circ\)
- \(60^\circ\)
- \(45^\circ\)
- \(30^\circ\)
Explanation:
समीकरण \(\dfrac{A\tan 62^\circ\sec 28^\circ\cot 38^\circ}{\csc 62^\circ\tan 11^\circ} = 1\) सही करने के लिए \(A\) का निम्नलिखित में से कौन-सा मान उपयुक्त होगा?
- \(\dfrac{\tan 38^\circ}{\tan 79^\circ\tan 28^\circ}\)
- \(\dfrac{\tan 28^\circ\tan 79^\circ}{\tan 38^\circ}\)
- \(\dfrac{\tan 28^\circ\tan 38^\circ}{\tan 79^\circ}\)
- \(\dfrac{\tan 38^\circ\tan 79^\circ}{\tan 28^\circ}\)
Explanation:
यदि \(a\sin A + b\cos A = c\) है, तो \(a\cos A – b\sin A\) का मान क्या है?
- \(\sqrt{a^2 – b^2 – c^2}\)
- \(\sqrt{a^2 + b^2 – c^2}\)
- \(\sqrt{a^2 + b^2 + c^2}\)
- \(\sqrt{a^2 – b^2 + c^2}\)
Explanation:
यदि \(X = \tan 40^\circ\) है, तो \(2\tan 50^\circ\) का मान ज्ञात करें।
- \(\dfrac{1}{2X}\)
- \(2X\)
- \(\dfrac{1}{X}\)
- \(\dfrac{2}{X}\)
Explanation:
यदि \(\dfrac{\cos A}{\csc A+1} + \dfrac{\cos A}{\csc A-1} = 2,\ 0^\circ \le A \le 90^\circ\) है, तो A का मान बताएं।
- \(60^\circ\)
- \(45^\circ\)
- \(90^\circ\)
- \(30^\circ\)
Explanation:
\(\dfrac{\tan 50^\circ + \sec 50^\circ}{\cot 40^\circ + \csc 40^\circ} + \cos^2 65^\circ + \sin 65^\circ\cos 25^\circ + \tan 30^\circ = ?\)
- \(\dfrac{\sqrt{3}(\sqrt{3}+1)}{3}\)
- \(2+\sqrt{3}\)
- \(\dfrac{6+\sqrt{3}}{3}\)
- \(1+\sqrt{3}\)
Explanation:
यदि \(\dfrac{\cot\theta+\cos\theta}{\cot\theta-\cos\theta} = \dfrac{k+1}{1-k}\), है, तो k का मान के बराबर है।
- \(\sin\theta\)
- \(\csc\theta\)
- \(\cos\theta\)
- \(\sec\theta\)
Explanation:
निम्न समीकरण में \(\theta\) (डिग्री में) का सही मान क्या होगा? \(\sin 3\theta\cos\theta – \cos 3\theta\sin\theta = \dfrac{1}{2},\ 0 < \theta < \dfrac{\pi}{2}\).
- \(45^\circ\)
- \(30^\circ\)
- \(60^\circ\)
- \(15^\circ\)
Explanation:
\(\theta\) के लिए : \(0^\circ < \theta < 90^\circ\) है। \(3\sec\theta + 4\cos\theta = 4\sqrt{3}\) है, तो \((1 – \sin\theta + \cos\theta)\) का मान ज्ञात करें।
- \(\dfrac{1+2\sqrt{3}}{2}\)
- \(\dfrac{1+\sqrt{3}}{2}\)
- \(\dfrac{1-\sqrt{3}}{2}\)
- \(\dfrac{1-2\sqrt{3}}{2}\)
Explanation:
यदि \(\dfrac{5\cot\theta+\sqrt{3}\csc\theta}{2\sqrt{3}\csc\theta+3\cot\theta} = 1,\ 0^\circ < \theta < 90^\circ\), तो \(\dfrac{\frac{7}{2}\cot^2\theta – \frac{3}{4}\csc^2\theta}{4\sin^2\theta + \frac{3}{2}\tan^2\theta}\) मान ज्ञात करें।
- \(7\)
- \(2\)
- \(3\)
- \(5\)
Explanation:
यदि \(\cot A + \csc A = 2\) है और \(A\) न्यून कोण है, तो \(\dfrac{9\tan A + 16\csc A}{5\sin A + 3\tan A}\) का मान ज्ञात करें।
- \(3\)
- \(8\)
- \(4\)
- \(8\)
Explanation:
यदि \(\sin\alpha + \csc\alpha = \tan\dfrac{\pi}{3}\) है, तो \((\sin^3\alpha + \csc^3\alpha)\) का मान ज्ञात कीजिए।
- \(\dfrac{1}{2}\)
- \(1\)
- \(0\)
- \(\dfrac{3}{2}\)
Explanation:
यदि \(0^\circ < \theta < 90^\circ\) है, तो \(\dfrac{(1-\sin\theta)(\csc\theta+\tan\theta)\tan\theta}{(\tan\theta+\sec\theta)(\cot\theta-\csc\theta+1)}\) का मान ज्ञात करें।
- \(\sin\theta\)
- \(\dfrac{\sin\theta}{2}\)
- \(\cos\theta\)
- \(\dfrac{\cos\theta}{2}\)
Explanation:
यदि \(\dfrac{3\sqrt{3}\sec\theta + 4\tan\theta}{3\tan\theta + \sqrt{3}\sec\theta} = 2,\ 0^\circ < \theta < 90^\circ\) है, तो \(\cos\theta\) का मान ज्ञात करें।
- \(\dfrac{\sqrt{3}}{2}\)
- \(\dfrac{1}{\sqrt{2}}\)
- \(\dfrac{1}{4}\)
- \(\dfrac{1}{2}\)
Explanation:
निम्नलिखित समीकरण को हल करें। \(\theta:\ 2\cos^2\theta + (4+\sqrt{3})\sin\theta – 2(1+\sqrt{3}) = 0\) जहां \(\theta\) एक न्यून कोण है।
- \(30^\circ\)
- \(45^\circ\)
- \(15^\circ\)
- \(60^\circ\)
Explanation:
यदि \((\cos^2\theta – 1)(2\sec^2\theta) + \sec^2\theta + 2\tan^2\theta = 2,\ 0^\circ < \theta < 90^\circ\) है, तो \(\dfrac{(\sec\theta+\sin\theta)}{(\csc\theta-\cos\theta)}\) का मान ज्ञात करें।
- \(-1\)
- \(-3\)
- \(3\)
- \(2\)
Explanation:
\(\dfrac{(1+\cos\theta)(\csc\theta-\cot\theta)\sec\theta}{\sin\theta(1-\sin\theta)(\sec\theta+\tan\theta)} = ?\)
- \(\sec^2\theta\)
- \(\sin^2\theta\)
- \(\cos^2\theta\)
- \(\csc^2\theta\)
Explanation:
\((1 + \tan 10^\circ)(1 + \tan 35^\circ) = ?\)
- \(\dfrac{1}{2}\)
- \(\dfrac{3}{4}\)
- \(1\)
- \(2\)
Explanation:
\(\dfrac{\sin 4\theta + \sin 2\theta}{\cos 4\theta + \cos 2\theta} = ?\)
- \(\tan 3\theta\)
- \(\cot 3\theta\)
- \(\tan\theta\)
- \(\cot\theta\)
Explanation:
\(\sin 18^\circ\) का मान \(\dfrac{\sqrt{5}-1}{4}\) के रूप में दिया गया है। इस मान का उपयोग करते हुए \(\cos 18^\circ\) का मान ज्ञात करें।
- \(\dfrac{\sqrt{10+2\sqrt{5}}}{4}\)
- \(\dfrac{\sqrt{5}+1}{4}\)
- \(\dfrac{\sqrt{6}+1}{4}\)
- \(\dfrac{\sqrt{10-2\sqrt{5}}}{4}\)
Explanation:
\(2\cos A\cos B = \cos(A+B) + \cos(A-B)\) के उपयोग से, \(\cos 75^\circ\cos 15^\circ\) का मान ज्ञात कीजिए।
- \(\dfrac{1}{2}\)
- \(\dfrac{1}{4}\)
- \(\dfrac{\sqrt{3}}{2}\)
- \(\dfrac{\sqrt{3}}{4}\)
Explanation:
\(\dfrac{\sin 4\theta}{1-\cos 4\theta} = ?\)
- \(\cot\theta\)
- \(\cot 2\theta\)
- \(\tan\theta\)
- \(\tan 2\theta\)
Explanation:
व्यंजक \(100(\sin 15^\circ\cos 15^\circ)\) का मान क्या है?
- \(50\)
- \(75\)
- \(100\)
- \(25\)
Explanation:
\(\sin 18^\circ\) का मान \(\dfrac{\sqrt{5}-1}{4}\) के रूप में दिया गया है। \(\csc 18^\circ\) का मान ज्ञात कीजिए।
- \(\dfrac{\sqrt{5}+1}{4}\)
- \(\dfrac{\sqrt{5}+1}{2}\)
- \(\sqrt{5}-1\)
- \(\sqrt{5}+1\)
Explanation:
\(\tan\dfrac{x}{2} = \dfrac{1-\cos x}{\sin x}\) सूत्र का उपयोग करते हुए \(\tan 22.5^\circ\) का मान ज्ञात कीजिए।
- \(\sqrt{2}-1\)
- \(\sqrt{2}+1\)
- \(\dfrac{\sqrt{3}+\sqrt{2}}{2}\)
- \(\dfrac{\sqrt{3}-\sqrt{2}}{2}\)
Explanation:
\(\sin 73^\circ + \cos 137^\circ\) का मान क्या होगा?
- \(\sin 13^\circ\)
- \(\cos 13^\circ\)
- \(\cos 18^\circ\)
- \(\sin 18^\circ\)
Explanation:
यदि \(\sin m + \sin n = p,\ \cos m + \cos n = q\), तो \(\sin m\times\sin n + \cos m\times\cos n\) का मान ज्ञात कीजिए।
- \(p^2 + q^2 – 2\)
- \(\dfrac{p^2+q^2-2}{2}\)
- \(p + q – pq\)
- \(p + q + pq\)
Explanation:
\(\dfrac{\cot\theta+\csc\theta-1}{\cot\theta-\csc\theta+1} = ?\)
- \(2\sec\theta\)
- \(2\csc\theta\)
- \(2\cot\theta\)
- \((\csc\theta + \cot\theta)\)
Explanation:
\(\dfrac{\tan\theta-\sec\theta+1}{\tan\theta+\sec\theta-1} = ?\)
- \(\dfrac{1+\sin\theta}{\cos\theta}\)
- \(\sec\theta + \tan\theta\)
- \(2\sec\theta\)
- \(\dfrac{\cos\theta}{1+\sin\theta}\)
Explanation:
\(\left(\dfrac{1+\sec^2 A}{1+\cos^2 A}\right)\left(\dfrac{1+\sin^2 A}{1+\csc^2 A}\right) = ?\)
- \(\cos^2 A\)
- \(\csc^2 A\)
- \(\sec^2 A\)
- \(\tan^2 A\)
Explanation:
\(\dfrac{\tan A-\sin A}{\tan A+\sin A} = ?\)
- \(1\)
- \(\dfrac{\sec A+1}{\sec A-1}\)
- \(\dfrac{\sec A-1}{\sec A+1}\)
- \(\sec A\)
Explanation:
\(\dfrac{\sqrt{(1-\sin A)}}{\sqrt{(1+\sin A)}} = ?\)
- \(\sec^2 A\)
- \(\sec A + \tan A\)
- \(\tan^2 A\)
- \(\sec A – \tan A\)
Explanation:
\(\sqrt{\dfrac{\csc A+1}{\csc A-1}} + \sqrt{\dfrac{\csc A-1}{\csc A+1}} = ?\)
- \(2\cos A\)
- \(\sec A\)
- \(\cos A\)
- \(2\sec A\)
Explanation:
\((\sec A – \tan A + 1)(\sec A – \tan A – 1) = ?\)
- \((1 + \tan A)\)
- \(2\tan A(\tan A – \sec A)\)
- \(2\tan A\)
- \(2\sec A\)
Explanation:
यदि \(-\sin\theta + \csc\theta = 6\) है, तो \(\sin\theta + \csc\theta\) का मान क्या है?
- \(6\)
- \(\sqrt{40}\)
- \(\sqrt{34}\)
- \(\sqrt{38}\)
Explanation:
\(\dfrac{1+\tan A}{\csc A} + \dfrac{1+\cot A}{\sec A}\)
- \(2\sec^2 A\)
- \(\sec A + \csc A\)
- \(\sec A + \csc A\)
- \(2\csc^2 A\)
Explanation:
व्यंजक \(4(\sin^6 A + \cos^6 A) – 6(\sin^4 A + \cos^4 A) + 8\) का मान ज्ञात कीजिए।
- \(8\)
- \(4\)
- \(6\)
- \(7\)
Explanation:
यदि \(\dfrac{\cos\theta}{1+\sin\theta} + \dfrac{\cos\theta}{1-\sin\theta} = 4\) और \(\theta\) न्यून है, तो \(\theta\) का मान क्या होगा?
- \(15^\circ\)
- \(30^\circ\)
- \(45^\circ\)
- \(60^\circ\)
Explanation:
\(\cos^4 A – \sin^4 A = ?\)
- \(2\sin^2 A – 1\)
- \(2\cos^2 A + 1\)
- \(1 – 2\sin^2 A\)
- \(-(2\sin^2 A + 1)\)
Explanation:
यदि \(a = \cot A + \cos A\) and \(b = \cot A – \cos A\), तो \(a^2 – b^2 – 4\sqrt{ab}\) का मान ज्ञात करें?
- \(0\)
- \(1\)
- \(-1\)
- \(-4\)
Explanation:
यदि \(2\tan^2 A + 4\cos^4 A = 3\) है, तो A का मान ज्ञात करें?
- \(45^\circ\)
- \(60^\circ\)
- \(0^\circ\)
- \(30^\circ\)
Explanation:
यदि \(\tan 15^\circ = 2 – \sqrt{3}\) है, तो \(\tan 15^\circ\cot 75^\circ + \tan 75^\circ\cot 15^\circ\) का मान ज्ञात कीजिए।
- \(6\)
- \(8\)
- \(14\)
- \(10\)
Explanation:
\((\cot B – \tan B)\cdot\sin B\cdot\cos B = ?\)
- \(1 – 2\cos^2 B\)
- \(2\cos^2 B – 1\)
- \(1 – 2\sec^2 B\)
- \(2\sec^2 B – 1\)
Explanation:
यदि \(\csc A + \cot A = a\sqrt{b}\) है, तो \(\dfrac{(a^2b-1)}{(a^2b+1)}\) का मान ज्ञात करें?
- \(\dfrac{1}{\cot A}\)
- \(\dfrac{1}{\sin A}\)
- \(\cos A\)
- \(\tan A\)
Explanation:
\(\dfrac{3(\sin^4 Z – \cos^4 Z + 1)}{\sin^2 Z} = ?\)
- \(9\)
- \(6\)
- \(2\)
- \(4\)
Explanation:
\((1 + \tan^2 A)(1 + \cot^2 A) = ?\)
- \(\dfrac{1}{\cos^2 A(1+\sin^2 A)}\)
- \(\dfrac{1}{\sin^2 A + \csc^2 A}\)
- \(\dfrac{1}{\sin^2 A(1-\sin^2 A)}\)
- \(\dfrac{1}{\sin^2 A(1+\cos^2 A)}\)
Explanation:
\(\cos(x+y) – \sin(x-y) + \tan(2z)\) का मान क्या है?
- \(\cos x\cos y – \sin x\sin y – \sin x\cos y – \cos x\sin y + \dfrac{2\tan z}{1-\tan^2 z}\)
- \(\cos x\cos y – \sin x\sin y – \sin x\cos y + \cos x\sin y + \dfrac{2\tan z}{1-\tan^2 z}\)
- \(\cos x\cos y – \sin x\sin y – \cos x\sin y + \dfrac{2\tan z}{1-\tan^2 z}\)
- \(\sin x\sin y + \dfrac{2\tan z}{1-\tan^2 z}\)
Explanation:
\(\left(\dfrac{\sin\theta}{1+\cos\theta} + \dfrac{1+\cos\theta}{\sin\theta}\right)\left(\dfrac{1}{\tan\theta+\cot\theta}\right)\) का मान किसके बराबर है?
- \(\tan\theta\)
- \(1\)
- \(2\cos\theta\)
- \(2\sin\theta\)
Explanation:
यदि \(\cot A = 7\), तो \(\dfrac{5\cos A + 4\sin A}{\cos^3 A + 7\sin^3 A + 6\sin A}\) का मान ज्ञात करें।
- \(3\)
- \(2\)
- \(1\)
- \(0\)
Explanation:
\(\dfrac{1}{\sin^4(90^\circ-2\alpha)} + \dfrac{1}{\cos^2(90^\circ-2\alpha)} – 1\) का मान क्या है?
- \(\sin^2 2\alpha + \tan^2 2\alpha\)
- \(\sec^2 2\alpha\times\tan^2 2\alpha\)
- \(\sec^2 2\alpha + \cot^2 2\alpha\)
- \(\cos^2 2\alpha\cot^2 2\alpha\)
Explanation:
यदि \(\sin(X – Y) = \sin X\cos Y – \cos X\sin Y\), तो \(\dfrac{\sin(A-B)}{\cos A\cos B} + \dfrac{\sin(B-C)}{\cos B\cos C}\) का मान ज्ञात कीजिए।
- \(\dfrac{\sin(C-A)}{\cos A\cos C}\)
- \(\dfrac{\sin(A-C)}{\cos A\cos C}\)
- \(\dfrac{\cos(A+C)}{\cos A\cos C}\)
- \(\dfrac{\cos(A+C)}{\sin A\sin C}\)
Explanation:
यदि \(\dfrac{\sec\theta+\tan\theta}{\sec\theta-\tan\theta} = \dfrac{5}{3},\ 0^\circ < \theta < 90^\circ\) है, तो \((\csc\theta + \cos\theta + \cot\theta)\) का मान क्या होगा?
- \(\dfrac{4+5\sqrt{15}}{2}\)
- \(\dfrac{8+3\sqrt{15}}{4}\)
- \(\dfrac{8+5\sqrt{15}}{4}\)
- \(\dfrac{16+5\sqrt{15}}{4}\)
Explanation:
\(\triangle ABC\) में, \(B\) पर समकोण है, यदि \(\cot C = 3\) है, तो \(\dfrac{\cos C(\sin C+\sin A)}{\sin A(\cos C-\cos A)}\) का मान इनमें से किसके बराबर होगा?
- \(3\)
- \(\dfrac{1}{2}\)
- \(1\)
- \(2\)
Explanation:
\(\sqrt{\dfrac{\cot\theta+\cos\theta}{\cot\theta-\cos\theta}}\) निम्न में से किसके बराबर है?
- \(1 + \sec\theta\tan\theta\)
- \(\sec\theta – \tan\theta\)
- \(\sec\theta + \tan\theta\)
- \(1 – \sec\theta – \tan\theta\)
Explanation:
यदि \(\theta\) प्रथम चतुर्थांश में है तथा \(\cos^2\theta – \sin^2\theta = \dfrac{1}{2}\) है, तो \(\tan^2 2\theta + \sin 3\theta\) का मान है
- \(4\)
- \(\dfrac{4}{3}\)
- \(\dfrac{7}{2}\)
- \(3\)
Explanation:
\(\dfrac{(1+\cos\theta)^2+\sin^2\theta}{(\csc^2\theta-1)\sin^2\theta} = ?\)
- \(\cos\theta(1+\sin\theta)\)
- \(2\sec\theta(1+\sec\theta)\)
- \(\sec\theta(1+\sin\theta)\)
- \(2\cos\theta(1+\sec\theta)\)
Explanation:
\(\dfrac{(2\sin A)(1+\sin A)}{1+\sin A+\cos A}\) का मान निम्न में से किसके बराबर है?
- \(1 + \sin A\cos A\)
- \(1 – \sin A\cos A\)
- \(1 + \cos A – \sin A\)
- \(1 + \sin A – \cos A\)
Explanation:
व्यंजक \((\cos^6\theta + \sin^6\theta – 1)(\tan^2\theta + \cot^2\theta + 2)\) निम्न में से किसके बराबर है?
- \(-3\)
- \(0\)
- \(1\)
- \(-1\)
Explanation:
\(\sqrt{\dfrac{\csc\theta-\cot\theta}{\csc\theta+\cot\theta}} \div \dfrac{\sin\theta}{1+\cos\theta} = ?\)
- \(\csc\theta\)
- \(\dfrac{1}{2}\)
- \(\sec\theta\)
- \(1\)
Explanation:
\((1 + \cot\theta – \csc\theta)(1 + \cos\theta + \sin\theta)\cdot\sec\theta = ?\)
- \(-2\)
- \(2\)
- \(\sec\theta\cdot\csc\theta\)
- \(\sin\theta\cdot\cos\theta\)
Explanation:
\(\dfrac{2(\sin^6\theta+\cos^6\theta)-3(\sin^4\theta+\cos^4\theta)}{\cos^4\theta-\sin^4\theta-2\cos^2\theta} = ?\)
- \(-1\)
- \(-2\)
- \(2\)
- \(1\)
Explanation:
\(\dfrac{(\sin\theta-\cos\theta)(1+\tan\theta+\cot\theta)}{1+\sin\theta\cos\theta} = ?\)
- \(\sec\theta – \csc\theta\)
- \(\csc\theta – \sec\theta\)
- \(\sin\theta + \cos\theta\)
- \(\tan\theta – \cot\theta\)
Explanation:
\(\dfrac{\sin\theta+\cos\theta-1}{\sin\theta-\cos\theta+1}\times\dfrac{\tan^2\theta(\csc^2\theta-1)}{\sec\theta-\tan\theta} = ?\)
- \(0\)
- \(-1\)
- \(1\)
- \(\dfrac{1}{2}\)
Explanation:
\(\dfrac{\sin\theta[(1-\tan\theta)\tan\theta+\sec^2\theta]}{(1-\sin\theta)\tan\theta(1+\tan\theta)(\sec\theta+\tan\theta)} = ?\)
- \(1\)
- \(\csc\theta\cdot\sec\theta\)
- \(\sin\theta\cdot\cos\theta\)
- \(-1\)
Explanation:
\(\left(\dfrac{1}{\cos\theta}-\dfrac{1}{\sin\theta}\right)+\dfrac{1}{\csc\theta-\cot\theta}-\dfrac{1}{\sec\theta+\tan\theta} = ?\)
- \(\sec\theta\cdot\csc\theta\)
- \(\sin\theta\cdot\tan\theta\)
- \(\csc\theta\cdot\cot\theta\)
- \(\sin\theta\cdot\cos\theta\)
Explanation:
\(\dfrac{\cos^6\theta+\sin^6\theta+3\sin^2\theta\cos^2\theta}{\csc\theta\sec\theta(\sin\theta+\cos\theta-1)(\sin\theta+\cos\theta+1)}\) का मान होगा:
- \(1\)
- \(2\)
- \(1/2\)
- \(3\)
Explanation:
\(\triangle ABC\) में, जो \(B\) पर समकोण है, यदि \(\tan A = 1/2\) है, तो \(\dfrac{\sin A(\cos C+\cos A)}{\cos C(\sin C-\sin A)}\) का मान ज्ञात कीजिए।
- \(2\sqrt{5}\)
- \(3\)
- \(2\)
- \(1\)
Explanation:
\(\dfrac{\sin\theta+\cos\theta-1}{\sin\theta-\cos\theta+1}\times\sqrt{\dfrac{1+\sin\theta}{1-\sin\theta}} = ?\)
- \(1\)
- \(-1\)
- \(-2\)
- \(2\)
Explanation:
\(\dfrac{(1+\tan\theta+\sec\theta)(1+\cot\theta-\csc\theta)}{(\sec\theta+\tan\theta)(1-\sin\theta)}\)
- \(2\sec\theta\)
- \(2\sec\theta\)
- \(2\csc\theta\)
- \(\csc\theta\)
Explanation:
\(\dfrac{\sec A(\sec A+\tan A)(1-\sin A)}{(\csc^2 A-1)\sin^2 A} = ?\)
- \(\sec^2 A\)
- \(\cos^2 A\)
- \(\cot A\)
- \(\cos A\)
Explanation:
यदि \(3\sin x + 4\cos x = 2\) है, तो \(3\cos x – 4\sin x\) का मान ज्ञात कीजिए।
- \(\sqrt{23}\)
- \(\sqrt{29}\)
- \(21\)
- \(\sqrt{21}\)
Explanation:
\(\left(\dfrac{\tan^3\theta}{\sec^2\theta}+\dfrac{\cot^3\theta}{\csc^2\theta}+2\sin\theta\cos\theta\right)\div(1+\csc^2\theta+\tan^2\theta),\ 0^\circ < \theta < 90^\circ\) का मान इनमें से किसके बराबर होगा?
- \(\sin\theta\cdot\cos\theta\)
- \(\csc\theta\)
- \(\sec\theta\)
- \(\csc\theta\cdot\sec\theta\)
Explanation:
माना कि \(0^\circ < \theta < 90^\circ\), है तो \((1+\cot^2\theta)(1+\tan^2\theta)\times(\sin\theta-\csc\theta)(\cos\theta-\sec\theta)\) का मान इनमें से किसके बराबर होगा?
- \(\sec\theta\csc\theta\)
- \(\sec\theta + \csc\theta\)
- \(\sin\theta + \cos\theta\)
- \(\sin\theta\cos\theta\)
Explanation:
यदि \(0^\circ < \theta < 90^\circ\) तो \(\dfrac{(1+\sec\theta\csc\theta)^2(\sec\theta-\tan\theta)^2(1+\sin\theta)}{(\sin\theta+\sec\theta)^2+(\cos\theta+\csc\theta)^2}\) का मान ज्ञात करें।
- \(1 – \cos\theta\)
- \(1 – \sin\theta\)
- \(\cos\theta\)
- \(\sin\theta\)
Explanation:
यदि \(0^\circ < \theta < 90^\circ\) तो \(\dfrac{1+\cos\theta-\sin^2\theta}{\sin\theta(1+\cos\theta)}\times\dfrac{\sqrt{\sec^2\theta+\csc^2\theta}}{\tan\theta+\cot\theta}\) का मान ज्ञात करें।
- \(\tan\theta\)
- \(\sec\theta\)
- \(\csc\theta\)
- \(\cot\theta\)
Explanation:
The expression \((\tan\theta+\cot\theta)(\sec\theta+\tan\theta)(1-\sin\theta)\), \(0^\circ < \theta < 90^\circ\), is equal to:
- \(\sec\theta\)
- \(\csc\theta\)
- \(\cot\theta\)
- \(\sin\theta\)
Explanation:
व्यंजक \(\dfrac{\tan^6\theta-\sec^6\theta+3\sec^2\theta\tan^2\theta}{\tan^2\theta+\cot^2\theta+1}\) का मान बताइए, जहां \(0^\circ < \theta < 90^\circ\) है।
- \(\sec^2\theta\csc^2\theta\)
- \(-\sec^2\theta\csc^2\theta\)
- \(\cos^2\theta\sin^2\theta\)
- \(-\cos^2\theta\sin^2\theta\)
Explanation:
व्यंजक \(\dfrac{\cos^4\theta-\sin^4\theta+2\sin^2\theta+3}{(\csc\theta+\cot\theta+1)(\csc\theta-\cot\theta+1)-2}\) का मान बताइए, जहाँ \(0^\circ < \theta < 90^\circ\) है।
- \(\dfrac{1}{2}\sin\theta\)
- \(2\sin\theta\)
- \(\sec\theta\)
- \(2\csc\theta\)
Explanation:
व्यंजक \(\dfrac{(1-\sin\theta+\cos\theta)^2(1-\cos\theta)\sec^3\theta\csc^2\theta}{(\sec\theta-\tan\theta)(\tan\theta+\cot\theta)}\) का मान बताइए, जहां \(0^\circ < \theta < 90^\circ\) है।
- \(\sin\theta\)
- \(2\cos\theta\)
- \(\cot\theta\)
- \(2\tan\theta\)
Explanation:
व्यंजक \(\dfrac{(1-2\sin^2\theta\cos\theta)(\cot\theta+1)\cos\theta}{(\sin^4\theta+\cos^4\theta)(1+\tan\theta)\csc\theta}-1\) का मान बताइए, जहाँ \(0^\circ < \theta < 90^\circ\) है।
- \(\cos^2\theta\)
- \(-\sin^2\theta\)
- \(-\cos 2\theta\)
- \(\sin^2\theta\)
Explanation:
\(\left[\dfrac{\tan\theta+\sec\theta-1}{\tan\theta-\sec\theta+1}\right] = ?\)
- \(\dfrac{1+\sin\theta}{\cos\theta}\)
- \(\dfrac{1+\tan\theta}{\cot\theta}\)
- \(\dfrac{1+\cot\theta}{\tan\theta}\)
- \(\dfrac{1+\cos\theta}{\sin\theta}\)
Explanation:
\((1 + \cot A + \tan A)(\sin A – \cos A)\left(\dfrac{\sin A\cos A}{\sin^3 A – \cos^3 A}\right) = ?\)
- \(1\)
- \(-1\)
- \(0\)
- \(2\)
Explanation:
यदि \(\sin x + \sin^2 x = 1\) है, तो \(\cos^8 x + 2\cos^6 x + \cos^4 x\) का मान क्या है?
- \(2\)
- \(0\)
- \(4\)
- \(1\)
Explanation:
यदि \(x\) वास्तविक राशि है, तो \((25\cos^2 x + 9\sec^2 x)\) का न्यूनतम मान ज्ञात करें।
- \(30\)
- \(20\)
- \(15\)
- \(40\)
Explanation:
\(\theta\) के सभी वास्तविक मानों के लिए \((\sin^{12}\theta + \cos^{20}\theta)\) का अधिकतम मान \(\_\_\_\_\_\_\) है।
- \(2\)
- \(3\)
- \(1\)
- \(0\)
Explanation:
\((2\sin\theta + 3\cos\theta)\) का अधिकतम मान है
- \(\sqrt{17}\)
- \(\sqrt{11}\)
- \(9\)
- \(\sqrt{13}\)
Explanation:
\(7\cos A + 24\sin A + 32\) का अधिकतम मान क्या है?
- \(57\)
- \(32\)
- \(25\)
- \(39\)
Explanation:
यदि \(x\) एक वास्तविक राशि है, तो \((16\cos^2 x + 9\sec^2 x)\) का न्यूनतम मान क्या है?
- \(24\)
- \(20\)
- \(30\)
- \(40\)
Explanation:
यह देखते हुए कि A और B दूसरे चतुर्थांश कोण हैं, \(\sin A = \dfrac{1}{3}\) और \(\sin B = \dfrac{1}{5}\) है, तो \(\cos(A-B)\) का मान ज्ञात कीजिए।
- \(\dfrac{4\sqrt{3}+1}{15}\)
- \(\dfrac{8\sqrt{3}-1}{15}\)
- \(\dfrac{8\sqrt{3}+1}{15}\)
- \(\dfrac{4\sqrt{3}+1}{15}\)
Explanation:
यदि \(\sin(A+B) = 1\) और \(\cos(A-B) = \dfrac{\sqrt{3}}{2}\) है, \(A + B \le 90^\circ\) और \(A > B\), तो \(\dfrac{5\sin^2 B + 4\tan^2 A}{2\sin B\cos A}\) का मान ज्ञात करें।
- \(20\)
- \(26\tfrac{1}{2}\)
- \(18\)
- \(16\tfrac{1}{2}\)
Explanation:
यदि \(\cos(A-B) = \dfrac{\sqrt{3}}{2}\) और \(\sec A = 2\) है, \(0^\circ \le A, B \le 90^\circ\), तो B का माप क्या है?
- \(60^\circ\)
- \(0^\circ\)
- \(30^\circ\)
- \(90^\circ\)
Explanation:
यदि \(x + \dfrac{1}{x} = 2\cos\theta\) है, तो \(x^3 + \dfrac{1}{x^3} = ?\)
- \(2\cos 2\theta\)
- \(\cos 3\theta\)
- \(2\cos 3\theta\)
- \(\cos 2\theta\)
Explanation:
यदि \(\sin\theta + \cos\theta = \dfrac{\sqrt{3}-1}{2\sqrt{2}}\) है, तो \(\tan\theta + \cot\theta\) का मान क्या है?
- \(8(\sqrt{3}-2)\)
- \(12(\sqrt{3}-2)\)
- \(8(\sqrt{3}+2)\)
- \(12(\sqrt{3}+2)\)
Explanation:
यदि \(\tan(\alpha+\beta) = \sqrt{3}\), \(\tan(\alpha-\beta) = 1\) जहां पर \((\alpha+\beta)\) तथा \((\alpha-\beta)\) न्यून कोण हैं, तो \(\tan(6\alpha)\) का मान क्या है?
- \(-1\)
- \(0\)
- \(1\)
- \(3\)
Explanation:
यदि \(P\cos\alpha = 3\) और \(4\tan\alpha = Q\) है, तो P और Q के बीच का संबंध है, जो \(\alpha\) से स्वतंत्र है?
- \(\dfrac{9}{P^2}+\dfrac{16}{Q^2} = 1\)
- \(\dfrac{9}{P^2}-\dfrac{16}{Q^2} = 1\)
- \(\dfrac{P^2}{9}-\dfrac{Q^2}{16} = 1\)
- \(\dfrac{P^2}{9}+\dfrac{Q^2}{16} = 1\)
Explanation:
\(\tan 5^\circ\tan 25^\circ\tan 45^\circ\tan 65^\circ\tan 85^\circ\) का मान \(\_\_\_\_\_\_\) के बराबर है।
- \(4\)
- \(3\)
- \(1\)
- \(2\)
Explanation:
यदि \(r = 15(\sin\theta+\cos\theta)\) और \(s = 16(\sin\theta-\cos\theta)\), तो \(\dfrac{r^2}{15^2}+\dfrac{s^2}{16^2}\) का मान ज्ञात कीजिए।
- \(8\)
- \(6\)
- \(4\)
- \(2\)
Explanation:
यदि \(x = a\cos\theta + b\sin\theta\) और \(y = a\sin\theta – b\cos\theta\), \(x^2 + y^2\) का मान है:
- \(a^2 – b^2\)
- \(a – b\)
- \(a^2 + b^2\)
- \(a + b\)
Explanation:
यदि \(\dfrac{\cos\theta}{1-\sin\theta}+\dfrac{\cos\theta}{1+\sin\theta} = 4,\ 0^\circ < \theta < 90^\circ\), तो \((\sec\theta + \csc\theta + \cot\theta)\) का मान ज्ञात कीजिए।
- \(1 + 2\sqrt{3}\)
- \(\dfrac{2+\sqrt{3}}{3}\)
- \(2 + \sqrt{3}\)
- \(\dfrac{1+2\sqrt{3}}{3}\)
Explanation:
\((1 + \sin^4 A – \cos^4 A)\csc^2 A = ?\)
- \(-2\)
- \(2\)
- \(1\)
- \(-1\)
Explanation:
\(\dfrac{[(\sin x+\sin y)(\sin x-\sin y)]}{[(\cos x+\cos y)(\cos y-\cos x)]} = ?\)
- \(0\)
- \(1\)
- \(-1\)
- \(2\)
Explanation:
निम्न में से कौन सा सत्य है जब \(x = \sin A + \cos A;\ y = \sec A + \csc A\)?
- \(y(1 + x^2) = 2x\)
- \(y – 2x = x^2 y\)
- \(y + 2x = x^2 y\)
- \(y(1 – 2x^2) = x\)
Explanation:
यदि \(\sec\theta = 4x\) और \(\tan\theta = \dfrac{4}{x},\ (x \ne 0)\) है, तो \(8\left(x^2 – \dfrac{1}{x^2}\right)\) का मान ज्ञात कीजिए।
- \(\dfrac{1}{16}\)
- \(\dfrac{1}{4}\)
- \(\dfrac{1}{2}\)
- \(\dfrac{1}{8}\)
Explanation:
यदि \(\csc\theta = 3x\) और \(\cot\theta = \dfrac{3}{x},\ (x \ne 0)\) है, तो \(6\left(x^2 – \dfrac{1}{x^2}\right)\) का मान है?
- \(\dfrac{2}{3}\)
- \(1\)
- \(\dfrac{1}{4}\)
- \(\dfrac{1}{2}\)
Explanation:
यदि \(\sin\theta = 3x\) और \(\cos\theta = \dfrac{3}{x},\ (x \ne 0)\) है, तो \(6\left(x^2 + \dfrac{1}{x^2}\right)\) का मान ज्ञात करें।
- \(\dfrac{1}{4}\)
- \(\dfrac{1}{3}\)
- \(\dfrac{2}{3}\)
- \(\dfrac{1}{2}\)
Explanation:
\(\left[\dfrac{\sqrt{3}+2\sin P}{1-2\cos P}\right]^3 + \left[\dfrac{1+2\cos P}{\sqrt{3}-2\sin P}\right]^3 = ?\)
- \(0\)
- \(\sin P\cos P\)
- \(2\sin P\cos P\)
- \(1\)
Explanation:
\(\sec\theta\) और \(\sin\theta\) (\(0^\circ < \theta < 90^\circ\)), \(\sqrt{6}x^2 – kx + \sqrt{6} = 0\) समीकरण के मूल हैं तो k का मान है:
- \(\sqrt{3}\)
- \(3\sqrt{2}\)
- \(2\sqrt{3}\)
- \(3\sqrt{3}\)
Explanation:
यदि \(\sin x = \dfrac{2}{3}\) है, तो \(\cos 3x\) का मान ज्ञात कीजिए।
- \(0.6735\)
- \(-0.8765\)
- \(-0.5797\)
- \(0.5678\)
Explanation:
यदि \(\tan\alpha = \dfrac{2}{\sqrt{13}}\) है, तो \(\dfrac{\csc^2\alpha+2\sec^2\alpha}{\csc^2\alpha-3\sec^2\alpha}\) का मान ज्ञात कीजिए।
- \(21\)
- \(14\)
- \(32\)
- \(16\)
Explanation:
सबसे छोटा धनात्मक कोण ज्ञात कीजिए जो दिए गए त्रिकोणमितीय समीकरण को संतुष्ट करता है। \(2\sin^2 x + \sqrt{3}\cos x + 1 = 0\)
- \(\dfrac{\pi}{3}\)
- \(\dfrac{2\pi}{3}\)
- \(\dfrac{5\pi}{6}\)
- \(\dfrac{\pi}{6}\)
Explanation:
\(\theta\) से \(x\cos\theta – y\sin\theta = 2\) और \(x\sin\theta + y\cos\theta = 4\) से \(\theta\) का विलोपन प्राप्त करेगा:
- \(3x^2 – y^2 = 20\)
- \(x^2 + y^2 = 20\)
- \(3x^2 + y^2 = 20\)
- \(x^2 – y^2 = 20\)
Explanation:
यदि \(\tan\theta = \dfrac{2}{\sqrt{11}},\ 0 < \theta < 90^\circ\), तो \(\dfrac{2\csc^2\theta-3\sec^2\theta}{3\csc^2\theta+4\sec^2\theta}\) का मान बराबर है:
- \(\dfrac{11}{45}\)
- \(\dfrac{11}{49}\)
- \(\dfrac{13}{49}\)
- \(\dfrac{10}{49}\)
Explanation:
\(\cot^2 46^\circ – \sec^2 44^\circ + (\sin^2 1^\circ + \sin^2 3^\circ + \sin^2 5^\circ + \ldots + \sin^2 89^\circ)\) का मान ज्ञात करें।
- \(23.5\)
- \(20.5\)
- \(22.5\)
- \(21.5\)
Explanation:
यदि \((1+\cot^2\theta)+(1+(\cot^2\theta)^{-1})\), k के बराबर है, तो \(\sqrt{k}\) का मान ज्ञात करें।
- \(\sin\theta\sec\theta\)
- \(\csc\theta\cos\theta\)
- \(\csc\theta\sec\theta\)
- \(\sin\theta\cos\theta\)
Explanation:
यदि \(5k = \tan\theta\) और \(\dfrac{5}{k} = \sec\theta\) है, तो \(10\left(k^2 – \dfrac{1}{k^2}\right)\) का मान क्या है?
- \(\dfrac{2}{5}\)
- \(-2\)
- \(-\dfrac{2}{5}\)
- \(2\)
Explanation:
यदि \(\cot\theta = \cot 30^\circ\cot 60^\circ\) और \(\theta\) कोई न्यून कोण है, तो \(2\theta\) \(\_\_\_\_\_\) के बराबर है।
- \(90^\circ\)
- \(60^\circ\)
- \(45^\circ\)
- \(30^\circ\)
Explanation:
यदि \(\sec x + \tan x = 5\) और \(\csc y – \cot y = \dfrac{1}{3}\) है, तो \((\sec x + \csc y) – (\tan x – \cot y)\) का मान ज्ञात करें।
- \(2.2\)
- \(3.2\)
- \(4.2\)
- \(3.1\)
Explanation:
यदि \(A = a\csc\theta + b\cot\theta\) और \(B = a\cot\theta + b\csc\theta\) है, तो \(A^2 – B^2\) का मान क्या है?
- \(ab\)
- \((a+b)\)
- \((a^2 – b^2)\)
- \((a-b)\)
Explanation:
\([\cot 1^\circ\cdot\cot 2^\circ\cdot\cot 3^\circ\cdot\cot 4^\circ\cdot\cot 5^\circ\ldots\cot 178^\circ\cdot\cot 179^\circ] = ?\)
- \(1235\)
- \(0\)
- \(1\)
- \(\dfrac{1}{2}\)
Explanation:
यदि \(\tan A = \dfrac{1}{2}\) और \(\tan B = \dfrac{1}{3}\) है, तो \(A + B\) का मान ज्ञात करें?
- \(45^\circ\)
- \(60^\circ\)
- \(30^\circ\)
- \(15^\circ\)
Explanation:
यदि \(\tan A = \sqrt{2} – 1\), तो \(\sin A\cos A + \tan A\) का मान ज्ञात कीजिये।
- \(\dfrac{24\sqrt{2}+15}{4}\)
- \(\dfrac{5\sqrt{2}-4}{4}\)
- \(\dfrac{24\sqrt{2}-15}{4}\)
- \(\dfrac{5\sqrt{2}+4}{4}\)
Explanation:
यदि \(\cot\alpha = \dfrac{P}{Q}\), तो \(\dfrac{P\cos\alpha-Q\sin\alpha}{P\cos\alpha+Q\sin\alpha} – \dfrac{P^2-Q^2}{P^2+Q^2} + 3\) का मान ज्ञात कीजिये।
- \(\dfrac{P^2}{P^2+Q^2}\)
- \(\dfrac{P}{Q}\)
- \(0\)
- \(3\)
Explanation:
यदि \(\tan A + \sin A = x,\ \tan A – \sin A = y\), तो \(x^2 – y^2\) का मान क्या है?
- \(2\sqrt{\dfrac{x}{y}}\)
- \(4\sqrt{\dfrac{x}{y}}\)
- \(\sqrt{xy}\)
- \(4\sqrt{xy}\)
Explanation:
यदि \(\alpha + \beta = 45^\circ\) और \((\tan\alpha + 1)(\tan\beta + 1) = 2x\), तो \(x\) का मान ज्ञात कीजिये।
- \(2\)
- \(-1\)
- \(0\)
- \(1\)
Explanation:
यदि \(f(p) = \sin^{p+2}x + \cos^{p+2}x\), तो \(6f(2) – 4f(4) + 10f(0)\) का मान ज्ञात कीजिये।
- \(14\)
- \(10\)
- \(12\)
- \(11\)
Explanation:
\(\cos^{-1}\left(\dfrac{7}{25}\right) – \cos^{-1}\left(\dfrac{15}{17}\right)\) का मान क्या है?
- \(\sin^{-1}\dfrac{304}{425}\)
- \(\cot^{-1}\dfrac{297}{425}\)
- \(\cos^{-1}\dfrac{297}{425}\)
- \(\cos^{-1}\dfrac{304}{425}\)
Explanation:
\(\dfrac{\tan 25^\circ\cot 65^\circ-\csc^2 65^\circ+\cot^2 61^\circ-\sec^2 29^\circ}{\sin^2 5^\circ+\sin^2 7^\circ+\ldots+\sin^2 85^\circ} = ?\)
- \(\dfrac{2}{45}\)
- \(\dfrac{4}{45}\)
- \(\dfrac{-2}{41}\)
- \(\dfrac{-4}{41}\)
Explanation:
यदि \(\sec\theta = \dfrac{a}{b},\ b \ne 0\) है, तो \(\dfrac{1-\tan^2\theta}{2-\sin^2\theta} = ?\)
- \(\dfrac{a^2(2b^2+a^2)}{b^2(a^2-b^2)}\)
- \(\dfrac{a^2(2b^2+a^2)}{b^2(a^2+b^2)}\)
- \(\dfrac{a^2(2b^2-a^2)}{b^2(a^2+b^2)}\)
- \(\dfrac{b^2(2b^2-a^2)}{a^2(a^2+b^2)}\)
Explanation:
यदि \(x = \sec 57^\circ\) है, तो \(\cot^2 33^\circ + \sin^2 57^\circ + \sin^2 33^\circ + \csc^2 57^\circ\cos^2 33^\circ + \sec^2 33^\circ\sin^2 57^\circ\), के बराबर हैं।
- \(x^2 + 2\)
- \(2x^2 + 1\)
- \(x^2 + 1\)
- \(\dfrac{1}{x^2+1}\)
Explanation:
यदि \(\csc\theta = \dfrac{b}{a}\) है, तो \(\dfrac{\sqrt{3}\cot\theta+1}{\tan\theta+\sqrt{3}}\) \(\ldots\ldots\) के बराबर है।
- \(\dfrac{\sqrt{b^2-a^2}}{a}\)
- \(\dfrac{\sqrt{a^2+b^2}}{a}\)
- \(\dfrac{\sqrt{a^2+b^2}}{b}\)
- \(\dfrac{\sqrt{b^2-a^2}}{b}\)
Explanation:
यदि \(\alpha + \beta = 90^\circ\) और \(\alpha = 2\beta\) है, तो \(3\cos^2\alpha – 2\sin^2\beta\) का मान ज्ञात कीजिए।
- \(\dfrac{4}{3}\)
- \(\dfrac{3}{4}\)
- \(\dfrac{3}{2}\)
- \(\dfrac{1}{4}\)
Explanation:
यदि \(\tan(A+B) = \sqrt{3}\) और \(\tan(A-B) = \dfrac{1}{\sqrt{3}};\ 0^\circ < (A+B) < 90^\circ;\ A > B\), तो A और B के मान क्रमशः \(\_\_\_\_\) हैं।
- \(45^\circ\ \&\ 15^\circ\)
- \(15^\circ\ \&\ 45^\circ\)
- \(30^\circ\ \&\ 30^\circ\)
- \(60^\circ\ \&\ 30^\circ\)
Explanation: