maths-trigonometry Mixed 1 – SARKARI LIBRARY

Mixed|MATHS

यदि \(\sec 4\theta = \operatorname{cosec}(\theta + 20^\circ)\) है, तो \(\theta\) किसके बराबर होगा?

\(22^\circ\)
\(18^\circ\)
\(14^\circ\)
\(20^\circ\)

Explanation:

\(\sec 4\theta = \operatorname{cosec}(\theta+20^\circ)\)
\(\sec x = \operatorname{cosec}(90^\circ – x)\) के अनुसार \(4\theta + (\theta+20^\circ) = 90^\circ\)
\(5\theta = 70^\circ \Rightarrow \theta = 14^\circ\)

यदि \(\tan 4\theta = \cot(2\theta + 30^\circ)\) है, तो \(\theta\) किसके बराबर होगा?

\(15^\circ\)
\(10^\circ\)
\(20^\circ\)
\(25^\circ\)

Explanation:

\(\tan 4\theta = \cot(2\theta+30^\circ)\)
\(\tan x = \cot(90^\circ – x)\) के अनुसार \(4\theta + 2\theta + 30^\circ = 90^\circ\)
\(6\theta = 60^\circ \Rightarrow \theta = 10^\circ\)

यदि \(\sin\theta = \cos(50^\circ + \theta)\), तो \(\theta\) का मान है:

\(20^\circ\)
\(25^\circ\)
\(30^\circ\)
\(35^\circ\)

Explanation:

\(\sin\theta = \cos(50^\circ+\theta)\)
\(\sin\theta = \cos(90^\circ-\theta)\) के अनुसार \(90^\circ – \theta = 50^\circ + \theta\)
\(2\theta = 40^\circ \Rightarrow \theta = 20^\circ\)

यदि \(\operatorname{cosec}4\theta = \sec(60^\circ – 2\theta)\) है तो \(\theta\) बराबर है:

\(18^\circ\)
\(25^\circ\)
\(15^\circ\)
\(20^\circ\)

Explanation:

\(\operatorname{cosec}4\theta = \sec(60^\circ-2\theta)\)
\(\operatorname{cosec}x = \sec(90^\circ-x)\) के अनुसार \(4\theta + 60^\circ – 2\theta = 90^\circ\)
\(2\theta = 30^\circ \Rightarrow \theta = 15^\circ\)

यदि \((20 + x)^\circ = \cos 60^\circ\), \(0 \le (20 + x) \le 90\) है, तो \(2\sin^2(3x + 15)^\circ – \operatorname{cosec}^2(2x + 10)^\circ\) का मान ज्ञात करें।

\(3\)
\(-3\)
\(-2\)
\(-\frac{1}{3}\)

Explanation:

प्रश्न में “\((20+x)^\circ = \cos 60^\circ\)” का आशय \(\sin(20+x)^\circ = \cos 60^\circ\) है, क्योंकि \(\cos 60^\circ = \sin 30^\circ\)
\(20 + x = 30 \Rightarrow x = 10\)
\(3x+15 = 45\) और \(2x+10 = 30\)
\(2\sin^2 45^\circ – \operatorname{cosec}^2 30^\circ = 2\times\dfrac{1}{2} – 4 = -3\)
नोट: प्रश्न में \(\sin\) छूटा हुआ प्रतीत होता है; इसी के अनुसार हल करने पर कुंजी का उत्तर \(-3\) मिलता है।

यदि \(\sec(5\alpha – 15^\circ) = \operatorname{cosec}(15^\circ – 2\alpha)\) है, तो \(\cos\alpha + \sin 2\alpha + \tan(1.5\alpha)\) का मान ज्ञात करें।

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

Explanation:

\(5\alpha – 15^\circ + 15^\circ – 2\alpha = 90^\circ \Rightarrow 3\alpha = 90^\circ \Rightarrow \alpha = 30^\circ\)
\(\cos 30^\circ + \sin 60^\circ + \tan 45^\circ = \dfrac{\sqrt{3}}{2} + \dfrac{\sqrt{3}}{2} + 1 = \sqrt{3}+1\)

यदि \(\cos(2\theta + 54^\circ) = \sin\theta\), \(0^\circ < (2\theta + 54^\circ) < 90^\circ\) है, तो \(\dfrac{1}{\cot 5\theta + \sec\dfrac{5\theta}{2}}\) का मान ज्ञात करें।

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

Explanation:

\(\cos(2\theta+54^\circ) = \sin\theta = \cos(90^\circ – \theta)\)
\(2\theta + 54^\circ = 90^\circ – \theta \Rightarrow \theta = 12^\circ\)
\(5\theta = 60^\circ,\ \dfrac{5\theta}{2} = 30^\circ\)
\(\cot 60^\circ + \sec 30^\circ = \dfrac{1}{\sqrt{3}} + \dfrac{2}{\sqrt{3}} = \sqrt{3}\)
मान \(= \dfrac{1}{\sqrt{3}} = \dfrac{\sqrt{3}}{3}\)

यदि \(\operatorname{cosec}A = \sec B\) है, जहाँ A और B न्यून कोण हैं, तो \((A + B)\) का मान क्या है?

\(0^\circ\)
\(135^\circ\)
\(90^\circ\)
\(45^\circ\)

Explanation:

\(\operatorname{cosec}A = \sec B\) और \(\operatorname{cosec}A = \sec(90^\circ – A)\)
अतः \(B = 90^\circ – A\) (न्यून कोण)
\(A + B = 90^\circ\)

यदि \(\sec 2\theta = \operatorname{cosec}(\theta – 36^\circ)\), जहां \(2\theta\) एक न्यून कोण है, तो \(\theta\) का मान ज्ञात कीजिए।

\(32^\circ\)
\(46^\circ\)
\(20^\circ\)
\(42^\circ\)

Explanation:

\(\sec 2\theta = \operatorname{cosec}(\theta – 36^\circ) \Rightarrow 2\theta + \theta – 36^\circ = 90^\circ\)
\(3\theta = 126^\circ \Rightarrow \theta = 42^\circ\)

यदि \(\sec 3\theta = \operatorname{cosec}(4\theta – 15^\circ)\) है, तो \(\tan 3\theta\) का मान ज्ञात कीजिए।

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

Explanation:

\(3\theta + 4\theta – 15^\circ = 90^\circ \Rightarrow 7\theta = 105^\circ \Rightarrow \theta = 15^\circ\)
\(\tan 3\theta = \tan 45^\circ = 1\)

यदि \(\sin(5x – 25^\circ) = \cos(5y + 25^\circ)\) है, जहां \(5x – 25^\circ\) और \(5y + 25^\circ\) न्यूनकोण है, तो \((x + y)\) का मान ज्ञात कीजिए।

\(50^\circ\)
\(40^\circ\)
\(18^\circ\)
\(16^\circ\)

Explanation:

\(\sin(5x-25^\circ) = \cos(5y+25^\circ) \Rightarrow (5x-25^\circ) + (5y+25^\circ) = 90^\circ\)
\(5(x+y) = 90^\circ \Rightarrow x+y = 18^\circ\)

\(\left[\dfrac{\sin^2 24^\circ + \sin^2 66^\circ}{\cos^2 24^\circ + \cos^2 66^\circ} + \sin^2 61^\circ + \cos 61^\circ \sin 29^\circ\right]\) का मान है:

\(1\)
\(3\)
\(0\)
\(2\)

Explanation:

\(\sin 66^\circ = \cos 24^\circ\) और \(\cos 66^\circ = \sin 24^\circ\)
\(\dfrac{\sin^2 24^\circ + \cos^2 24^\circ}{\cos^2 24^\circ + \sin^2 24^\circ} = 1\)
\(\sin 29^\circ = \cos 61^\circ\), अतः \(\sin^2 61^\circ + \cos^2 61^\circ = 1\)
कुल मान \(= 1 + 1 = 2\)

यदि \(\sec 2x = \operatorname{cosec}(3x – 45^\circ)\) है, तब \(x\) का मान क्या है?

\(27^\circ\)
\(40^\circ\)
\(45^\circ\)
\(35^\circ\)

Explanation:

\(2x + 3x – 45^\circ = 90^\circ \Rightarrow 5x = 135^\circ \Rightarrow x = 27^\circ\)

यदि \(0 \le \theta \le 90^\circ\), और \(\sin(2\theta + 50^\circ) = \cos(4\theta + 16^\circ)\) है, तो \(\theta\) का मान (अंश में) ज्ञात कीजिए।

\(8^\circ\)
\(10^\circ\)
\(12^\circ\)
\(4^\circ\)

Explanation:

\(\sin(2\theta+50^\circ) = \cos(4\theta+16^\circ) \Rightarrow 2\theta + 50^\circ + 4\theta + 16^\circ = 90^\circ\)
\(6\theta = 24^\circ \Rightarrow \theta = 4^\circ\)

यदि \(\tan(11\theta) = \cot(7\theta)\) है, तो \(\sin^2(6\theta) + \sec^2(9\theta) + \operatorname{cosec}^2(12\theta)\) का मान ज्ञात कीजिए।

\(\dfrac{35}{12}\)
\(\dfrac{23}{6}\)
\(\dfrac{31}{12}\)
\(\dfrac{43}{12}\)

Explanation:

\(\tan 11\theta = \cot 7\theta \Rightarrow 11\theta + 7\theta = 90^\circ \Rightarrow \theta = 5^\circ\)
\(\sin^2 30^\circ + \sec^2 45^\circ + \operatorname{cosec}^2 60^\circ = \dfrac{1}{4} + 2 + \dfrac{4}{3}\)
\(= \dfrac{3 + 24 + 16}{12} = \dfrac{43}{12}\)

यदि \(\sec 3x = \operatorname{cosec}(3x – 45^\circ)\) है, जहां \(3x\), न्यून कोण है, तो \(x\) का मान ज्ञात कीजिए।

\(27.5^\circ\)
\(22.5^\circ\)
\(35^\circ\)
\(45^\circ\)

Explanation:

\(3x + 3x – 45^\circ = 90^\circ \Rightarrow 6x = 135^\circ \Rightarrow x = 22.5^\circ\)

यदि \(4\theta\) न्यून कोण है और \(\cot 4\theta = \tan(\theta – 5^\circ)\) है, तो \(\theta\) का मान ज्ञात करें।

\(19^\circ\)
\(45^\circ\)
\(21^\circ\)
\(24^\circ\)

Explanation:

\(\cot 4\theta = \tan(\theta – 5^\circ) \Rightarrow 4\theta + \theta – 5^\circ = 90^\circ\)
\(5\theta = 95^\circ \Rightarrow \theta = 19^\circ\)

यदि \(A + B = 90^\circ\) है, तो व्यंजक \(\dfrac{\cot A}{\cot B} + \cos^2 A + \cos^2 B\) बराबर है

\(\cot^2 B\)
\(\operatorname{cosec}^2 A\)
\(\cot^2 A\)
\(\operatorname{cosec}^2 B\)

Explanation:

\(A + B = 90^\circ \Rightarrow \cot B = \tan A,\ \cos B = \sin A\)
\(\dfrac{\cot A}{\cot B} = \dfrac{\cot A}{\tan A} = \cot^2 A\)
\(\cos^2 A + \cos^2 B = \cos^2 A + \sin^2 A = 1\)
व्यंजक \(= \cot^2 A + 1 = \operatorname{cosec}^2 A\)

\(\sin^2 15^\circ + \sin^2 25^\circ + \sin^2 35^\circ + \sin^2 45^\circ + \sin^2 55^\circ + \sin^2 65^\circ + \sin^2 75^\circ\) का मान है:

\(4\)
\(7\)
\(\dfrac{7}{2}\)
\(\dfrac{7}{3}\)

Explanation:

\(\sin 75^\circ = \cos 15^\circ,\ \sin 65^\circ = \cos 25^\circ,\ \sin 55^\circ = \cos 35^\circ\)
\((\sin^2 15^\circ + \cos^2 15^\circ) + (\sin^2 25^\circ + \cos^2 25^\circ) + (\sin^2 35^\circ + \cos^2 35^\circ) + \sin^2 45^\circ\)
\(= 1 + 1 + 1 + \dfrac{1}{2} = \dfrac{7}{2}\)

यदि \(x, y\) न्यून कोण हैं, जहां \(0 < x + y < 90^\circ\) और \(\sin(3x – 40^\circ) = \cos(3y + 40^\circ)\), तो \(\tan(x + y)\) का मान बराबर है:

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

Explanation:

\(\sin(3x-40^\circ) = \cos(3y+40^\circ) \Rightarrow 3x – 40^\circ + 3y + 40^\circ = 90^\circ\)
\(3(x+y) = 90^\circ \Rightarrow x+y = 30^\circ\)
\(\tan 30^\circ = \dfrac{1}{\sqrt{3}}\)

\(\operatorname{cosec}(58^\circ + \theta) – \sec(32^\circ – \theta) + \sin 15^\circ \sin 35^\circ \sec 55^\circ \sin 30^\circ \sec 75^\circ = ?\)

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

Explanation:

\((58^\circ+\theta) + (32^\circ-\theta) = 90^\circ\), अतः \(\operatorname{cosec}(58^\circ+\theta) = \sec(32^\circ-\theta)\) और इनका अंतर \(0\) है।
\(\sec 55^\circ = \dfrac{1}{\cos 55^\circ} = \dfrac{1}{\sin 35^\circ}\) और \(\sec 75^\circ = \dfrac{1}{\sin 15^\circ}\)
\(\sin 15^\circ \sin 35^\circ \cdot\dfrac{1}{\sin 35^\circ}\cdot\sin 30^\circ\cdot\dfrac{1}{\sin 15^\circ} = \sin 30^\circ = \dfrac{1}{2}\)
कुल मान \(= 0 + \dfrac{1}{2} = \dfrac{1}{2}\)

यदि \(\sin B – \cos A = 0\), तो \(A + B\) का मान क्या है?

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

Explanation:

\(\sin B = \cos A = \sin(90^\circ – A)\)
\(B = 90^\circ – A \Rightarrow A + B = 90^\circ\)

यदि \(\sec(5a – 15^\circ) = \operatorname{cosec}(15^\circ – 2a)\) है, तो \(\cos 3a + \sin 2a + \cot a\) का मान-

\(\dfrac{3\sqrt{3}}{2}\)
\(\dfrac{2}{\sqrt{3}}\)
\(\dfrac{\sqrt{17}}{2}\)
\(18\)

Explanation:

\(5a – 15^\circ + 15^\circ – 2a = 90^\circ \Rightarrow a = 30^\circ\)
\(\cos 90^\circ + \sin 60^\circ + \cot 30^\circ = 0 + \dfrac{\sqrt{3}}{2} + \sqrt{3} = \dfrac{3\sqrt{3}}{2}\)

\(\dfrac{(\cos 9^\circ + \sin 81^\circ)(\sec 9^\circ + \operatorname{cosec}81^\circ)}{\sin 56^\circ \sec 34^\circ + \cos 25^\circ \operatorname{cosec}65^\circ} = ?\)

\(0.5\)
\(4\)
\(2\)
\(0.25\)

Explanation:

\(\cos 9^\circ = \sin 81^\circ\) और \(\sec 9^\circ = \operatorname{cosec}81^\circ\)
अंश \(= (2\cos 9^\circ)(2\sec 9^\circ) = 4\)
\(\sin 56^\circ \sec 34^\circ = \dfrac{\sin 56^\circ}{\cos 34^\circ} = 1\) और \(\cos 25^\circ \operatorname{cosec}65^\circ = \dfrac{\cos 25^\circ}{\sin 65^\circ} = 1\)
हर \(= 2\)
मान \(= \dfrac{4}{2} = 2\)

\(\dfrac{\sin(78^\circ + \theta) – \cos(12^\circ – \theta) + (\tan^2 70^\circ – \operatorname{cosec}^2 20^\circ)}{\sin 25^\circ \cos 65^\circ + \cos 25^\circ \sin 65^\circ}\) का मान है:

\(2\)
\(-1\)
\(-2\)
\(0\)

Explanation:

\(\sin(78^\circ+\theta) = \cos(12^\circ-\theta)\), अतः पहले दो पद कटकर \(0\) हो जाते हैं।
\(\tan^2 70^\circ – \operatorname{cosec}^2 20^\circ = \tan^2 70^\circ – \sec^2 70^\circ = -1\)
हर: \(\sin 25^\circ \cos 65^\circ + \cos 25^\circ \sin 65^\circ = \sin^2 25^\circ + \cos^2 25^\circ = 1\)
मान \(= \dfrac{-1}{1} = -1\)

\((\tan 29^\circ \cot 61^\circ – \operatorname{cosec}^2 61^\circ) + \cot^2 54^\circ – \sec^2 36^\circ + (\sin^2 1^\circ + \sin^2 3^\circ + \sin^2 5^\circ + \dots + \sin^2 89^\circ)\) का मान है:

\(20\frac{1}{2}\)
\(21\)
\(22\frac{1}{2}\)
\(22\)

Explanation:

\(\cot 61^\circ = \tan 29^\circ\), अतः \(\tan 29^\circ \cot 61^\circ = \tan^2 29^\circ\) और \(\operatorname{cosec}^2 61^\circ = \sec^2 29^\circ\)
\(\tan^2 29^\circ – \sec^2 29^\circ = -1\)
\(\cot^2 54^\circ – \sec^2 36^\circ = \cot^2 54^\circ – \operatorname{cosec}^2 54^\circ = -1\)
\(\sin^2 1^\circ + \sin^2 3^\circ + \dots + \sin^2 89^\circ\) में 45 पद हैं; \(\sin^2 \theta + \sin^2(90^\circ-\theta) = 1\) से 22 जोड़े \(= 22\) और बीच का पद \(\sin^2 45^\circ = \dfrac{1}{2}\)
कुल \(= -1 – 1 + 22\dfrac{1}{2} = 20\dfrac{1}{2}\)

\(\operatorname{cosec}(67^\circ + \theta) – \sec(23^\circ – \theta) + \cos 15^\circ \cos 35^\circ \operatorname{cosec}55^\circ \cdot \cos 60^\circ \cdot \operatorname{cosec}75^\circ = ?\)

\(2\)
\(0\)
\(1\)
\(0.5\)

Explanation:

\((67^\circ+\theta) + (23^\circ-\theta) = 90^\circ\), अतः \(\operatorname{cosec}(67^\circ+\theta) – \sec(23^\circ-\theta) = 0\)
\(\operatorname{cosec}55^\circ = \dfrac{1}{\cos 35^\circ}\) और \(\operatorname{cosec}75^\circ = \dfrac{1}{\cos 15^\circ}\)
बाकी भाग \(= \cos 60^\circ = 0.5\)
मान \(= 0.5\)

\(\sin^2 64^\circ + \cos 64^\circ \sin 26^\circ + 2\cos 43^\circ \operatorname{cosec}47^\circ = ?\)

\(4\)
\(1\)
\(2\)
\(3\)

Explanation:

\(\sin 26^\circ = \cos 64^\circ\), अतः \(\sin^2 64^\circ + \cos 64^\circ \sin 26^\circ = \sin^2 64^\circ + \cos^2 64^\circ = 1\)
\(2\cos 43^\circ \operatorname{cosec}47^\circ = 2\dfrac{\cos 43^\circ}{\sin 47^\circ} = 2\)
कुल मान \(= 1 + 2 = 3\)

यदि \(\sin 3A = \cos(A + 10^\circ)\), जहां \(3A\) न्यून कोण है, तो \(2\operatorname{cosec}\dfrac{3A}{2} + 6\sin^2 3A – \dfrac{3}{2}\tan^2 3A\) का मान ज्ञात कीजिए।

\(\dfrac{7}{4}\)
\(4\)
\(\dfrac{17}{2}\)
\(5\)

Explanation:

\(\sin 3A = \cos(A+10^\circ) \Rightarrow 3A + A + 10^\circ = 90^\circ \Rightarrow A = 20^\circ\)
\(3A = 60^\circ,\ \dfrac{3A}{2} = 30^\circ\)
\(2\operatorname{cosec}30^\circ + 6\sin^2 60^\circ – \dfrac{3}{2}\tan^2 60^\circ = 4 + \dfrac{9}{2} – \dfrac{9}{2} = 4\)

यदि \(\cos(2\theta + 54^\circ) = \sin\theta\), \(0^\circ < (2\theta + 54^\circ) < 90^\circ\) है, तो \(\dfrac{1}{\tan 5\theta + \operatorname{cosec}\dfrac{5\theta}{2}}\) का मान ज्ञात कीजिए।

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

Explanation:

\(\cos(2\theta+54^\circ) = \sin\theta \Rightarrow 2\theta + 54^\circ = 90^\circ – \theta \Rightarrow \theta = 12^\circ\)
\(5\theta = 60^\circ,\ \dfrac{5\theta}{2} = 30^\circ\)
\(\tan 60^\circ + \operatorname{cosec}30^\circ = \sqrt{3} + 2\)
मान \(= \dfrac{1}{2+\sqrt{3}} = \dfrac{2-\sqrt{3}}{(2+\sqrt{3})(2-\sqrt{3})} = 2-\sqrt{3}\)

यदि \(\sin(x + y) = \cos(x – y)\) है, तो \(\cos^2 x\) का मान ज्ञात कीजिए।

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

Explanation:

\(\sin(x+y) = \cos(x-y)\) का विस्तार करने पर
\(\sin x\cos y + \cos x\sin y = \cos x\cos y + \sin x\sin y\)
\((\sin x – \cos x)(\cos y – \sin y) = 0\)
इसलिए \(\tan x = 1\) या \(\tan y = 1\); \(x = 45^\circ\) लेने पर \(\cos^2 45^\circ = \dfrac{1}{2}\)
नोट: यदि \(y = 45^\circ\) हो तो \(\cos^2 x\) निश्चित नहीं होता; प्रश्न का आशय \(x = 45^\circ\) है, जो कुंजी से मेल खाता है।

\(\dfrac{\cos 50^\circ}{\sin 40^\circ} + \dfrac{3\operatorname{cosec}80^\circ}{\sec 10^\circ} – 2\cos 50^\circ \cdot \operatorname{cosec}40^\circ = ?\)

\(3\)
\(2\)
\(5\)
\(4\)

Explanation:

\(\dfrac{\cos 50^\circ}{\sin 40^\circ} = 1\) (क्योंकि \(\cos 50^\circ = \sin 40^\circ\))
\(\dfrac{3\operatorname{cosec}80^\circ}{\sec 10^\circ} = \dfrac{3\cos 10^\circ}{\sin 80^\circ} = 3\)
\(2\cos 50^\circ \operatorname{cosec}40^\circ = 2\)
मान \(= 1 + 3 – 2 = 2\)

यदि \(\cot\theta = \sqrt{7}\) है, तो \(\dfrac{\operatorname{cosec}^2\theta – \sec^2\theta}{\operatorname{cosec}^2\theta + \sec^2\theta}\) का मान है:

\(\dfrac{3}{4}\)
\(\dfrac{2}{3}\)
\(\dfrac{8}{9}\)
\(\dfrac{7}{9}\)

Explanation:

\(\operatorname{cosec}^2\theta = 1 + \cot^2\theta = 8\)
\(\tan\theta = \dfrac{1}{\sqrt{7}} \Rightarrow \sec^2\theta = 1 + \dfrac{1}{7} = \dfrac{8}{7}\)
मान \(= \dfrac{8 – \frac{8}{7}}{8 + \frac{8}{7}} = \dfrac{48/7}{64/7} = \dfrac{3}{4}\)

यदि \(\tan\theta = \dfrac{2}{3}\), तो \(\dfrac{3\sin\theta – 4\cos\theta}{3\sin\theta + 4\cos\theta}\) किसके बराबर होगा?

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

Explanation:

अंश और हर को \(\cos\theta\) से भाग देने पर
\(\dfrac{3\tan\theta – 4}{3\tan\theta + 4} = \dfrac{3\times\frac{2}{3} – 4}{3\times\frac{2}{3} + 4} = \dfrac{2-4}{2+4} = -\dfrac{1}{3}\)

यदि \(\cos\theta = \dfrac{4}{5}\) है, तो \(\sin^2\theta \cdot \cos\theta + \cos^2\theta \cdot \sin\theta\) बराबर है:

\(\dfrac{16}{25}\)
\(\dfrac{93}{125}\)
\(\dfrac{14}{25}\)
\(\dfrac{84}{125}\)

Explanation:

\(\cos\theta = \dfrac{4}{5}\) (न्यून कोण मानने पर) \(\Rightarrow \sin\theta = \dfrac{3}{5}\)
\(\sin^2\theta\cos\theta + \cos^2\theta\sin\theta = \dfrac{9}{25}\times\dfrac{4}{5} + \dfrac{16}{25}\times\dfrac{3}{5}\)
\(= \dfrac{36}{125} + \dfrac{48}{125} = \dfrac{84}{125}\)

यदि \(3\sin\theta = 4\cos\theta\) है, तो \(\tan^2\theta + \sin\theta – \cos\theta\) बराबर है:

\(\dfrac{88}{45}\)
\(2\)
\(\dfrac{89}{45}\)
\(\dfrac{17}{9}\)

Explanation:

\(3\sin\theta = 4\cos\theta \Rightarrow \tan\theta = \dfrac{4}{3}\)
अतः \(\sin\theta = \dfrac{4}{5},\ \cos\theta = \dfrac{3}{5}\)
\(\tan^2\theta + \sin\theta – \cos\theta = \dfrac{16}{9} + \dfrac{4}{5} – \dfrac{3}{5} = \dfrac{16}{9} + \dfrac{1}{5}\)
\(= \dfrac{80 + 9}{45} = \dfrac{89}{45}\)

यदि \(2\sin\theta = 5\cos\theta\) है, तो \(\dfrac{\sin\theta + \cos\theta}{\sin\theta – \cos\theta}\) बराबर है:

\(5/3\)
\(9/5\)
\(2/3\)
\(7/3\)

Explanation:

\(2\sin\theta = 5\cos\theta \Rightarrow \tan\theta = \dfrac{5}{2}\)
अंश व हर को \(\cos\theta\) से भाग देने पर
\(\dfrac{\tan\theta + 1}{\tan\theta – 1} = \dfrac{\frac{5}{2}+1}{\frac{5}{2}-1} = \dfrac{7/2}{3/2} = \dfrac{7}{3}\)

यदि A प्रथम चतुर्थांश में अवस्थित है और \(6\tan A = 5\), तो \(\dfrac{8\sin A – 4\cos A}{\cos A + 2\sin A}\) का मान बताइए।

\(1\)
\(4\)
\(-2\)
\(16\)

Explanation:

\(6\tan A = 5 \Rightarrow \tan A = \dfrac{5}{6}\)
अंश व हर को \(\cos A\) से भाग देने पर
\(\dfrac{8\tan A – 4}{1 + 2\tan A} = \dfrac{\frac{20}{3} – 4}{1 + \frac{5}{3}} = \dfrac{8/3}{8/3} = 1\)

\(\operatorname{cosec}(85^\circ + \theta) – \sec(5^\circ – \theta) – \tan(55^\circ + \theta) + \cot(35^\circ – \theta) = ?\)

\(1\)
\(3/2\)
\(-1\)
\(0\)

Explanation:

\((85^\circ+\theta) + (5^\circ-\theta) = 90^\circ\), अतः \(\operatorname{cosec}(85^\circ+\theta) = \sec(5^\circ-\theta)\)
\((55^\circ+\theta) + (35^\circ-\theta) = 90^\circ\), अतः \(\tan(55^\circ+\theta) = \cot(35^\circ-\theta)\)
दोनों जोड़े कट जाते हैं, अतः मान \(= 0\)

यदि \(\dfrac{\sec\theta – \tan\theta}{\sec\theta + \tan\theta} = \dfrac{3}{5}\) है, तो \(\dfrac{\operatorname{cosec}\theta + \cot\theta}{\operatorname{cosec}\theta – \cot\theta}\) का मान बताइए।

\(31 + 8\sqrt{15}\)
\(33 + 4\sqrt{15}\)
\(27 + \sqrt{15}\)
\(24 + \sqrt{15}\)

Explanation:

\(\dfrac{\sec\theta – \tan\theta}{\sec\theta + \tan\theta} = \dfrac{1-\sin\theta}{1+\sin\theta} = \dfrac{3}{5}\)
\(5 – 5\sin\theta = 3 + 3\sin\theta \Rightarrow \sin\theta = \dfrac{1}{4}\)
\(\cos\theta = \dfrac{\sqrt{15}}{4}\)
\(\dfrac{\operatorname{cosec}\theta + \cot\theta}{\operatorname{cosec}\theta – \cot\theta} = \dfrac{1+\cos\theta}{1-\cos\theta} = \dfrac{4+\sqrt{15}}{4-\sqrt{15}}\)
\(= (4+\sqrt{15})^2 = 31 + 8\sqrt{15}\)

चित्र में, \(\cot\theta\) का मान क्या है? (समकोण त्रिभुज PQR में Q पर समकोण है, \(PQ = 8\), \(PR = 17\) और R पर कोण \(\theta\) है।)

\(\dfrac{17}{18}\)
\(\dfrac{15}{17}\)
\(\dfrac{15}{8}\)
\(\dfrac{8}{15}\)

Explanation:

समकोण त्रिभुज PQR में \(PR = 17,\ PQ = 8\)
\(QR = \sqrt{17^2 – 8^2} = \sqrt{225} = 15\)
R पर कोण \(\theta\) के लिए आधार \(QR = 15\) और लम्ब \(PQ = 8\)
\(\cot\theta = \dfrac{QR}{PQ} = \dfrac{15}{8}\)

यदि \(\sec\theta – \tan\theta = \dfrac{x}{y}\), \((0 < x < y)\) और \(0^\circ < \theta < 90^\circ\) है, तो \(\sin\theta\) का मान ज्ञात कीजिए।

\(\dfrac{x^2 + y^2}{2xy}\)
\(\dfrac{2xy}{x^2 + y^2}\)
\(\dfrac{y^2 – x^2}{x^2 + y^2}\)
\(\dfrac{x^2 + y^2}{y^2 – x^2}\)

Explanation:

\((\sec\theta – \tan\theta)(\sec\theta + \tan\theta) = 1 \Rightarrow \sec\theta + \tan\theta = \dfrac{y}{x}\)
\(\sec\theta = \dfrac{1}{2}\left(\dfrac{x}{y} + \dfrac{y}{x}\right) = \dfrac{x^2+y^2}{2xy}\)
\(\tan\theta = \dfrac{1}{2}\left(\dfrac{y}{x} – \dfrac{x}{y}\right) = \dfrac{y^2-x^2}{2xy}\)
\(\sin\theta = \dfrac{\tan\theta}{\sec\theta} = \dfrac{y^2-x^2}{x^2+y^2}\)

यदि \(5\sin\theta = 4\) है, तो \(\dfrac{\sec\theta + 4\cot\theta}{4\tan\theta – 5\cos\theta}\) का मान ज्ञात करें।

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

Explanation:

\(\sin\theta = \dfrac{4}{5} \Rightarrow \cos\theta = \dfrac{3}{5},\ \tan\theta = \dfrac{4}{3},\ \cot\theta = \dfrac{3}{4},\ \sec\theta = \dfrac{5}{3}\)
अंश \(= \dfrac{5}{3} + 3 = \dfrac{14}{3}\)
हर \(= \dfrac{16}{3} – 3 = \dfrac{7}{3}\)
मान \(= \dfrac{14/3}{7/3} = 2\)

यदि \(5\cos\theta – 12\sin\theta = 0\) है, तो \(\dfrac{1 + \sin\theta + \cos\theta}{1 – \sin\theta + \cos\theta}\) का मान क्या है?

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

Explanation:

\(5\cos\theta = 12\sin\theta \Rightarrow \tan\theta = \dfrac{5}{12}\)
अतः \(\sin\theta = \dfrac{5}{13},\ \cos\theta = \dfrac{12}{13}\)
\(\dfrac{1 + \frac{5}{13} + \frac{12}{13}}{1 – \frac{5}{13} + \frac{12}{13}} = \dfrac{30/13}{20/13} = \dfrac{3}{2}\)

यदि A तीसरे चतुर्थांश में स्थित है, और \(20\tan A = 21\) है, तो \(\dfrac{5\sin A – 2\cos A}{4\cos A – \frac{5}{7}\sin A}\) का मान ज्ञात कीजिए।

\(\dfrac{5}{29}\)
\(\dfrac{-65}{29}\)
\(1\)
\(\dfrac{13}{12}\)

Explanation:

\(\tan A = \dfrac{21}{20}\)
अंश व हर को \(\cos A\) से भाग देने पर
\(\dfrac{5\tan A – 2}{4 – \frac{5}{7}\tan A} = \dfrac{\frac{21}{4} – 2}{4 – \frac{3}{4}} = \dfrac{13/4}{13/4} = 1\)
(तीसरे चतुर्थांश में \(\sin A,\ \cos A\) दोनों ऋणात्मक हैं, पर अनुपात पर प्रभाव नहीं पड़ता।)

यदि \(\dfrac{\sin A + \cos A}{\cos A} = \dfrac{17}{12}\) है, तो \(\dfrac{1 – \cos A}{\sin A}\) का मान ज्ञात कीजिए।

\(1/5\)
\(5/12\)
\(-5\)
\(1\)

Explanation:

\(\dfrac{\sin A + \cos A}{\cos A} = \tan A + 1 = \dfrac{17}{12} \Rightarrow \tan A = \dfrac{5}{12}\)
\(\sin A = \dfrac{5}{13},\ \cos A = \dfrac{12}{13}\)
\(\dfrac{1 – \cos A}{\sin A} = \dfrac{1 – \frac{12}{13}}{\frac{5}{13}} = \dfrac{1}{5}\)

यदि \(\dfrac{\operatorname{cosec}\theta + \cot\theta}{\operatorname{cosec}\theta – \cot\theta} = 7\) है, तो \(\dfrac{4\sin^2\theta – 1}{4\sin^2\theta + 5}\) का मान ज्ञात करें।

\(\dfrac{1}{3}\)
\(-\dfrac{1}{3}\)
\(-\dfrac{1}{9}\)
\(\dfrac{1}{9}\)

Explanation:

\(\dfrac{\operatorname{cosec}\theta + \cot\theta}{\operatorname{cosec}\theta – \cot\theta} = \dfrac{1+\cos\theta}{1-\cos\theta} = 7\)
\(1 + \cos\theta = 7 – 7\cos\theta \Rightarrow \cos\theta = \dfrac{3}{4}\)
\(\sin^2\theta = 1 – \dfrac{9}{16} = \dfrac{7}{16} \Rightarrow 4\sin^2\theta = \dfrac{7}{4}\)
\(\dfrac{\frac{7}{4} – 1}{\frac{7}{4} + 5} = \dfrac{3/4}{27/4} = \dfrac{1}{9}\)

यदि \(2\sin(3x – 15)^\circ = 1\), \(0^\circ < (3x – 15)^\circ < 90^\circ\) है, तो \(\cos^2(2x + 15)^\circ + \cot^2(x + 15)^\circ\) का मान ज्ञात करें।

\(1\)
\(\dfrac{5}{2}\)
\(-\dfrac{7}{2}\)
\(\dfrac{7}{2}\)

Explanation:

\(2\sin(3x-15)^\circ = 1 \Rightarrow \sin(3x-15)^\circ = \dfrac{1}{2} = \sin 30^\circ\)
\(3x – 15 = 30 \Rightarrow x = 15\)
\(\cos^2(2x+15)^\circ + \cot^2(x+15)^\circ = \cos^2 45^\circ + \cot^2 30^\circ = \dfrac{1}{2} + 3 = \dfrac{7}{2}\)

यदि \(\sin A = \dfrac{1}{2}\), A न्यून कोण है, तो \(\dfrac{\tan A – \cot A}{\sqrt{3}(1 + \operatorname{cosec}A)}\) का मान ज्ञात करें।

\(-\dfrac{2}{9}\)
\(-\dfrac{4\sqrt{3}}{9}\)
\(\dfrac{4\sqrt{3}}{9}\)
\(\dfrac{2}{9}\)

Explanation:

\(\sin A = \dfrac{1}{2} \Rightarrow A = 30^\circ\)
\(\tan 30^\circ – \cot 30^\circ = \dfrac{1}{\sqrt{3}} – \sqrt{3} = -\dfrac{2}{\sqrt{3}}\)
\(\sqrt{3}(1 + \operatorname{cosec}30^\circ) = \sqrt{3}(1+2) = 3\sqrt{3}\)
मान \(= \dfrac{-2/\sqrt{3}}{3\sqrt{3}} = -\dfrac{2}{9}\)

यदि \(\tan\theta = \sqrt{5}\), तो \(\dfrac{\operatorname{cosec}^2\theta + \sec^2\theta}{\operatorname{cosec}^2\theta – \sec^2\theta}\) का मान ज्ञात करें।

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

Explanation:

\(\tan^2\theta = 5 \Rightarrow \operatorname{cosec}^2\theta = 1 + \cot^2\theta = 1 + \dfrac{1}{5} = \dfrac{6}{5}\)
\(\sec^2\theta = 1 + \tan^2\theta = 6\)
मान \(= \dfrac{\frac{6}{5} + 6}{\frac{6}{5} – 6} = \dfrac{36/5}{-24/5} = -\dfrac{3}{2}\)

\(\operatorname{cosec}(60^\circ + A) – \sec(30^\circ – A) + \dfrac{\operatorname{cosec}49^\circ}{\sec 41^\circ} = ?\)

\(2\)
\(-1\)
\(0\)
\(1\)

Explanation:

\((60^\circ + A) + (30^\circ – A) = 90^\circ\), अतः \(\operatorname{cosec}(60^\circ+A) = \sec(30^\circ-A)\) और इनका अंतर \(0\) है।
\(\dfrac{\operatorname{cosec}49^\circ}{\sec 41^\circ} = \dfrac{\cos 41^\circ}{\sin 49^\circ} = 1\) (क्योंकि \(\cos 41^\circ = \sin 49^\circ\))
मान \(= 0 + 1 = 1\)

यदि \(\sec 31^\circ = x\) है, तो \(\sin^2 59^\circ + \dfrac{1}{\sec^2 31^\circ} – \dfrac{1}{\sin^2 59^\circ \operatorname{cosec}^2 59^\circ}\) का मान ज्ञात करें।

\(\dfrac{x^2 – 2}{x}\)
\(\dfrac{2 – x^2}{x^2}\)
\(\dfrac{x^2 – 2}{x^2}\)
\(\dfrac{2 – x^2}{x}\)

Explanation:

\(\sin 59^\circ = \cos 31^\circ = \dfrac{1}{\sec 31^\circ} = \dfrac{1}{x}\)
\(\sin^2 59^\circ = \dfrac{1}{x^2}\) और \(\dfrac{1}{\sec^2 31^\circ} = \dfrac{1}{x^2}\)
\(\sin^2 59^\circ \operatorname{cosec}^2 59^\circ = 1\)
मान \(= \dfrac{1}{x^2} + \dfrac{1}{x^2} – 1 = \dfrac{2 – x^2}{x^2}\)

यदि \(\sec A = 17/8\) है, जबकि \(A < 90^\circ\) है, तो निम्नलिखित का मान क्या है? \(\dfrac{34\sin A + 15\cot A}{68\cos A – 16\tan A}\)

\(23\)
\(19\)
\(30\)
\(38\)

Explanation:

\(\sec A = \dfrac{17}{8} \Rightarrow \cos A = \dfrac{8}{17},\ \sin A = \dfrac{15}{17},\ \tan A = \dfrac{15}{8},\ \cot A = \dfrac{8}{15}\)
अंश \(= 34\times\dfrac{15}{17} + 15\times\dfrac{8}{15} = 30 + 8 = 38\)
हर \(= 68\times\dfrac{8}{17} – 16\times\dfrac{15}{8} = 32 – 30 = 2\)
मान \(= \dfrac{38}{2} = 19\)

यदि \(\cos B = \dfrac{5}{7}\) है, तो \(\operatorname{cosec}B + \cot B\) का मान ज्ञात करें। दिया गया कि \(0 < B < \dfrac{\pi}{2}\) है।

\(\dfrac{5}{\sqrt{6}}\)
\(\dfrac{\sqrt{6}}{12}\)
\(\dfrac{7}{\sqrt{6}}\)
\(\sqrt{6}\)

Explanation:

\(\cos B = \dfrac{5}{7} \Rightarrow \sin B = \sqrt{1 – \dfrac{25}{49}} = \dfrac{2\sqrt{6}}{7}\)
\(\operatorname{cosec}B + \cot B = \dfrac{1 + \cos B}{\sin B} = \dfrac{12/7}{2\sqrt{6}/7} = \dfrac{6}{\sqrt{6}} = \sqrt{6}\)

यदि \(\cos 53^\circ = \dfrac{x}{y}\) तब \(\sec 53^\circ + \cot 37^\circ\) का मान क्या होगा?

\(\dfrac{x + \sqrt{y^2 – x^2}}{y}\)
\(\dfrac{x + \sqrt{y^2 – x^2}}{x}\)
\(\dfrac{y + \sqrt{y^2 – x^2}}{x}\)
\(\dfrac{y + \sqrt{y^2 – x^2}}{y}\)

Explanation:

\(\sec 53^\circ = \dfrac{1}{\cos 53^\circ} = \dfrac{y}{x}\)
\(\sin 53^\circ = \dfrac{\sqrt{y^2 – x^2}}{y}\)
\(\cot 37^\circ = \tan 53^\circ = \dfrac{\sin 53^\circ}{\cos 53^\circ} = \dfrac{\sqrt{y^2 – x^2}}{x}\)
योग \(= \dfrac{y + \sqrt{y^2 – x^2}}{x}\)

यदि \(\tan B = \dfrac{5}{3}\) है, तो \(\dfrac{\operatorname{cosec}B + \sin B}{\cos B – \sec B}\) का मान क्या होगा?

\(-\dfrac{177}{125}\)
\(-\dfrac{59}{15}\)
\(\dfrac{59}{15}\)
\(\dfrac{177}{125}\)

Explanation:

\(\tan B = \dfrac{5}{3} \Rightarrow \sin B = \dfrac{5}{\sqrt{34}},\ \cos B = \dfrac{3}{\sqrt{34}}\)
अंश \(= \dfrac{\sqrt{34}}{5} + \dfrac{5}{\sqrt{34}} = \dfrac{34 + 25}{5\sqrt{34}} = \dfrac{59}{5\sqrt{34}}\)
हर \(= \dfrac{3}{\sqrt{34}} – \dfrac{\sqrt{34}}{3} = \dfrac{9 – 34}{3\sqrt{34}} = -\dfrac{25}{3\sqrt{34}}\)
मान \(= \dfrac{59}{5}\times\dfrac{3}{-25} = -\dfrac{177}{125}\)

यदि \(\sin\theta = \dfrac{1}{5}\) है, तो \(\dfrac{(1 + \cos\theta)}{\sin\theta}\) का मान क्या होगा?

\(5 + 2\sqrt{6}\)
\(\dfrac{2}{5}\)
\(\dfrac{4}{5}\)
\(\dfrac{1}{5}\)

Explanation:

\(\sin\theta = \dfrac{1}{5} \Rightarrow \cos\theta = \dfrac{\sqrt{24}}{5} = \dfrac{2\sqrt{6}}{5}\)
\(\dfrac{1 + \cos\theta}{\sin\theta} = \dfrac{1 + \frac{2\sqrt{6}}{5}}{\frac{1}{5}} = 5 + 2\sqrt{6}\)

यदि \(\sin A = \dfrac{4}{5}\) और \(\sin B = \dfrac{15}{17}\) है, तो \(\sin(A – B)\) का मान क्या होगा?

\(-\dfrac{32}{45}\)
\(-\dfrac{13}{85}\)
\(\dfrac{13}{85}\)
\(\dfrac{32}{45}\)

Explanation:

\(\sin A = \dfrac{4}{5} \Rightarrow \cos A = \dfrac{3}{5}\) और \(\sin B = \dfrac{15}{17} \Rightarrow \cos B = \dfrac{8}{17}\) (न्यून कोण)
\(\sin(A-B) = \sin A\cos B – \cos A\sin B\)
\(= \dfrac{4}{5}\times\dfrac{8}{17} – \dfrac{3}{5}\times\dfrac{15}{17} = \dfrac{32 – 45}{85} = -\dfrac{13}{85}\)

\(\dfrac{1}{1 + \cos(90^\circ – \theta)} + \dfrac{1}{1 – \cos(90^\circ – \theta)} = \_\_\_\_\_\_\_\_\)

\(2\sec^2\theta\)
\(1\)
\(0\)
\(2\tan^2\theta\)

Explanation:

\(\cos(90^\circ – \theta) = \sin\theta\)
\(\dfrac{1}{1+\sin\theta} + \dfrac{1}{1-\sin\theta} = \dfrac{2}{1 – \sin^2\theta} = \dfrac{2}{\cos^2\theta} = 2\sec^2\theta\)

\(\tan(\theta – 14\pi) = ?\)

\(\tan\theta\)
\(-\cot\theta\)
\(\cot\theta\)
\(-\tan\theta\)

Explanation:

\(\tan\) का आवर्त \(\pi\) है, अतः \(\tan(\theta – 14\pi) = \tan\theta\) (क्योंकि \(14\pi\) में \(\pi\) की पूर्ण संख्या है)।

यदि \(\cot B = 9\) है, तो \(\sec^2 B\) का मान क्या होगा?

\(\dfrac{98}{81}\)
\(\dfrac{92}{81}\)
\(\dfrac{42}{81}\)
\(\dfrac{82}{81}\)

Explanation:

\(\cot B = 9 \Rightarrow \tan B = \dfrac{1}{9}\)
\(\sec^2 B = 1 + \tan^2 B = 1 + \dfrac{1}{81} = \dfrac{82}{81}\)

यदि \(\tan A = \dfrac{2}{3}\) है, तो निम्नलिखित का मान क्या होगा? \((5\sin^2 A – 2\cos^2 A) \div (15\sin^2 A + 3\cos^2 A)\)

\(\dfrac{21}{47}\)
\(\dfrac{2}{87}\)
\(\dfrac{2}{5}\)
\(\dfrac{3}{77}\)

Explanation:

\(\tan A = \dfrac{2}{3}\); अंश व हर को \(\cos^2 A\) से भाग देने पर
\(\dfrac{5\tan^2 A – 2}{15\tan^2 A + 3} = \dfrac{5\times\frac{4}{9} – 2}{15\times\frac{4}{9} + 3} = \dfrac{2/9}{87/9} = \dfrac{2}{87}\)

यदि \(b\sin\theta = a\) है, तो \(\sec\theta + \tan\theta = ?\)

\(\sqrt{\dfrac{b + a}{b – a}}\)
\(\sqrt{\dfrac{1}{b + a}}\)
\(\sqrt{\dfrac{1}{b – a}}\)
\(\sqrt{\dfrac{b – a}{b + a}}\)

Explanation:

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

\(\dfrac{\cos 20^\circ}{\sin 70^\circ} + \dfrac{\cos\theta}{\sin(90^\circ – \theta)} = \_\_\_\_\_\_\_\_\)

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

Explanation:

\(\sin 70^\circ = \cos 20^\circ \Rightarrow \dfrac{\cos 20^\circ}{\sin 70^\circ} = 1\)
\(\sin(90^\circ – \theta) = \cos\theta \Rightarrow \dfrac{\cos\theta}{\sin(90^\circ – \theta)} = 1\)
योग \(= 2\)

\(\cos\left(-\dfrac{17\pi}{3}\right) = ?\)

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

Explanation:

\(\cos(-x) = \cos x\), अतः \(\cos\left(-\dfrac{17\pi}{3}\right) = \cos\dfrac{17\pi}{3}\)
\(\dfrac{17\pi}{3} = 6\pi – \dfrac{\pi}{3}\)
\(\cos\left(6\pi – \dfrac{\pi}{3}\right) = \cos\dfrac{\pi}{3} = \dfrac{1}{2}\)

\(\cos 15^\circ = ?\)

\(\sqrt{2 + \sqrt{3}}\)
\(\dfrac{\sqrt{2 + \sqrt{3}}}{4}\)
\(\dfrac{\sqrt{2 + \sqrt{3}}}{2}\)
\(\dfrac{1 + \sqrt{3}}{2}\)

Explanation:

\(\cos 15^\circ = \cos(45^\circ – 30^\circ) = \cos 45^\circ\cos 30^\circ + \sin 45^\circ\sin 30^\circ\)
\(= \dfrac{\sqrt{3}+1}{2\sqrt{2}} = \dfrac{\sqrt{6}+\sqrt{2}}{4}\)
वर्ग करने पर \(\cos^2 15^\circ = \dfrac{1 + \cos 30^\circ}{2} = \dfrac{2+\sqrt{3}}{4}\), अतः \(\cos 15^\circ = \dfrac{\sqrt{2+\sqrt{3}}}{2}\)
नोट: सही मान \(\dfrac{\sqrt{2+\sqrt{3}}}{2} \approx 0.9659\) है, जो विकल्प 3 है। कुंजी में विकल्प 2 (\(\frac{\sqrt{2+\sqrt{3}}}{4}\)) दिया है, जो सही नहीं है (उसका मान \(\approx 0.483\) आता है)। कृपया कुंजी जाँच लें।

\(\cos 45^\circ \sin 15^\circ\) का मान क्या है?

\(\dfrac{\sqrt{3} – 1}{2}\)
\(\dfrac{\sqrt{3} – 1}{4}\)
\(\dfrac{\sqrt{3} + 1}{2}\)
\(\dfrac{\sqrt{3} + 1}{4}\)

Explanation:

\(\cos 45^\circ\sin 15^\circ = \dfrac{1}{2}\left[\sin 60^\circ – \sin 30^\circ\right]\)
\(= \dfrac{1}{2}\left[\dfrac{\sqrt{3}}{2} – \dfrac{1}{2}\right] = \dfrac{\sqrt{3}-1}{4}\)

यदि \(4\tan\theta = 3\) है, तो \(\dfrac{4\sin\theta – \cos\theta + 1}{4\sin\theta + \cos\theta – 1}\) ________ होगा।

\(14/11\)
\(12/11\)
\(10/11\)
\(13/11\)

Explanation:

\(4\tan\theta = 3 \Rightarrow \tan\theta = \dfrac{3}{4} \Rightarrow \sin\theta = \dfrac{3}{5},\ \cos\theta = \dfrac{4}{5}\)
अंश \(= \dfrac{12}{5} – \dfrac{4}{5} + 1 = \dfrac{13}{5}\)
हर \(= \dfrac{12}{5} + \dfrac{4}{5} – 1 = \dfrac{11}{5}\)
मान \(= \dfrac{13}{11}\)

निम्नलिखित का मान ज्ञात कीजिए: \(\cos(36^\circ + A) \cdot \cos(36^\circ – A) + \cos(54^\circ + A) \cdot \cos(54^\circ – A)\)

\(\sin 2A\)
\(\cos A\)
\(\sin A\)
\(\cos 2A\)

Explanation:

\(\cos(36^\circ+A)\cos(36^\circ-A) = \dfrac{1}{2}\left[\cos 72^\circ + \cos 2A\right]\)
\(\cos(54^\circ+A)\cos(54^\circ-A) = \dfrac{1}{2}\left[\cos 108^\circ + \cos 2A\right]\)
जोड़ने पर \(= \dfrac{1}{2}\left[\cos 72^\circ + \cos 108^\circ\right] + \cos 2A\)
\(\cos 108^\circ = -\cos 72^\circ\), अतः मान \(= \cos 2A\)

यदि \(\sin 23^\circ = \dfrac{a}{b}\) है, तो \(\sin 23 – \sin 67^\circ\) का मान क्या है?

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

Explanation:

\(\sin 23^\circ = \dfrac{a}{b} \Rightarrow \cos 23^\circ = \dfrac{\sqrt{b^2 – a^2}}{b}\) और \(\sin 67^\circ = \cos 23^\circ\)
कुंजी के रूप \(\dfrac{a^2}{b\sqrt{b^2-a^2}}\) के लिए व्यंजक \(\sec 23^\circ – \sin 67^\circ\) लेना होगा:
\(\sec 23^\circ – \cos 23^\circ = \dfrac{1 – \cos^2 23^\circ}{\cos 23^\circ} = \dfrac{\sin^2 23^\circ}{\cos 23^\circ}\)
\(= \dfrac{a^2/b^2}{\sqrt{b^2-a^2}/b} = \dfrac{a^2}{b\sqrt{b^2 – a^2}}\)
नोट: प्रश्न में “\(\sin 23\)” की जगह “\(\sec 23^\circ\)” प्रतीत होता है। शाब्दिक \(\sin 23^\circ – \sin 67^\circ\) का मान \(\dfrac{a – \sqrt{b^2-a^2}}{b}\) होगा, जो किसी विकल्प से नहीं मिलता।

\(\cos 47^\circ \sec 133^\circ + \sin 44^\circ \operatorname{cosec}136^\circ\) का मान ज्ञात कीजिए:

\(1/2\)
\(1\)
\(0\)
\(-1\)

Explanation:

\(\sec 133^\circ = \dfrac{1}{\cos 133^\circ} = -\dfrac{1}{\cos 47^\circ}\) (क्योंकि \(\cos 133^\circ = -\cos 47^\circ\))
\(\cos 47^\circ \sec 133^\circ = -1\)
\(\operatorname{cosec}136^\circ = \dfrac{1}{\sin 136^\circ} = \dfrac{1}{\sin 44^\circ}\), अतः \(\sin 44^\circ\operatorname{cosec}136^\circ = 1\)
योग \(= -1 + 1 = 0\)

यदि \(\sin A = -\dfrac{3}{5}\) है, A तृतीय चतुर्थांश में स्थित है, तो \(\sec A\) का मान क्या है?

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

Explanation:

A तृतीय चतुर्थांश में है, अतः \(\cos A\) ऋणात्मक है।
\(\cos A = -\sqrt{1 – \dfrac{9}{25}} = -\dfrac{4}{5}\)
\(\sec A = -\dfrac{5}{4}\)

यदि \(\sin x = \dfrac{3}{10}\) है, तो \(\tan x + \cot x\) का मान क्या है?

\(\dfrac{100}{3\sqrt{91}}\)
\(\dfrac{100}{2\sqrt{83}}\)
\(\dfrac{100}{7\sqrt{95}}\)
\(\dfrac{100}{3\sqrt{85}}\)

Explanation:

\(\sin x = \dfrac{3}{10} \Rightarrow \cos x = \dfrac{\sqrt{91}}{10}\)
\(\tan x + \cot x = \dfrac{\sin^2 x + \cos^2 x}{\sin x\cos x} = \dfrac{1}{\sin x\cos x}\)
\(= \dfrac{1}{\frac{3}{10}\times\frac{\sqrt{91}}{10}} = \dfrac{100}{3\sqrt{91}}\)

\(\operatorname{cosec}2910^\circ + \sec4260^\circ + \tan2565^\circ + \cot1755^\circ = ?\)

\(3\)
\(1\)
\(4\)
\(0\)

Explanation:

\(2910^\circ = 8\times 360^\circ + 30^\circ \Rightarrow \operatorname{cosec}2910^\circ = \operatorname{cosec}30^\circ = 2\)
\(4260^\circ = 11\times 360^\circ + 300^\circ \Rightarrow \sec 300^\circ = \sec 60^\circ = 2\)
\(2565^\circ = 7\times 360^\circ + 45^\circ \Rightarrow \tan 45^\circ = 1\)
\(1755^\circ = 4\times 360^\circ + 315^\circ \Rightarrow \cot 315^\circ = -1\)
योग \(= 2 + 2 + 1 – 1 = 4\)

\(\sin(-405^\circ) = ?\)

\(\dfrac{\sqrt{5}}{2}\)
\(\dfrac{1}{2}\)
\(\dfrac{-1}{2}\)
\(\dfrac{-1}{\sqrt{2}}\)

Explanation:

\(\sin(-405^\circ) = -\sin 405^\circ = -\sin(360^\circ + 45^\circ) = -\sin 45^\circ = -\dfrac{1}{\sqrt{2}}\)

निम्नलिखित को हल कीजिए। \(2 + \cos 49^\circ \cos 41^\circ – \sin 49^\circ \sin 41^\circ\)

\(2\)
\(-1\)
\(0\)
\(1\)

Explanation:

\(\cos 49^\circ\cos 41^\circ – \sin 49^\circ\sin 41^\circ = \cos(49^\circ + 41^\circ) = \cos 90^\circ = 0\)
मान \(= 2 + 0 = 2\)

यदि \(\tan x = \dfrac{12}{5}\) है, जहां \(x\) दूसरे चतुर्थांश में स्थित है, तो \(\sin x – \cot x\) का मान कितना है?

\(\dfrac{209}{156}\)
\(\dfrac{169}{156}\)
\(\dfrac{1567}{209}\)
\(\dfrac{1448}{169}\)

Explanation:

\(\tan x = \dfrac{12}{5}\) और \(x\) दूसरे चतुर्थांश में है: \(\sin x\) धनात्मक, \(\cos x\) ऋणात्मक।
\(\sin x = \dfrac{12}{13},\ \cos x = -\dfrac{5}{13},\ \cot x = -\dfrac{5}{12}\)
\(\sin x – \cot x = \dfrac{12}{13} + \dfrac{5}{12} = \dfrac{144 + 65}{156} = \dfrac{209}{156}\)

यदि \(\tan x = \dfrac{7}{5}\) है, तो \(\dfrac{9\sin x + \frac{42}{5}\cos x}{15\sin x + 21\cos x}\) का मान ज्ञात कीजिए।

\(0\)
\(1\)
\(0.24\)
\(0.5\)

Explanation:

अंश व हर को \(\cos x\) से भाग देने पर
अंश \(= 9\tan x + \dfrac{42}{5} = 9\times\dfrac{7}{5} + \dfrac{42}{5} = \dfrac{63 + 42}{5} = 21\)
हर \(= 15\tan x + 21 = 21 + 21 = 42\)
मान \(= \dfrac{21}{42} = 0.5\)
नोट: सही मान \(0.5\) (विकल्प 4) आता है; कुंजी में \(0.24\) दिया है जो इस विधि से नहीं बनता। कृपया कुंजी जाँच लें।

\(\tan(-1125^\circ) = ?\)

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

Explanation:

\(\tan(-1125^\circ) = -\tan 1125^\circ\)
\(1125^\circ = 3\times 360^\circ + 45^\circ\)
\(-\tan 45^\circ = -1\)

\(\tan 4384^\circ + \cot 6814^\circ = ?\)

\(-1\)
\(2\)
\(0\)
\(1\)

Explanation:

\(4384^\circ = 12\times 360^\circ + 64^\circ \Rightarrow \tan 4384^\circ = \tan 64^\circ\)
\(6814^\circ = 18\times 360^\circ + 334^\circ \Rightarrow \cot 334^\circ = \cot(360^\circ – 26^\circ) = -\cot 26^\circ = -\tan 64^\circ\)
योग \(= \tan 64^\circ – \tan 64^\circ = 0\)

यदि \(4\tan\theta – 3 = 0\) है, तो \(\dfrac{1 – \cos 2\theta}{1 + \cos 2\theta}\) का मान ज्ञात कीजिए।

\(1\)
\(\dfrac{7}{15}\)
\(\dfrac{4}{3}\)
\(\dfrac{9}{16}\)

Explanation:

\(4\tan\theta – 3 = 0 \Rightarrow \tan\theta = \dfrac{3}{4}\)
\(\dfrac{1 – \cos 2\theta}{1 + \cos 2\theta} = \dfrac{2\sin^2\theta}{2\cos^2\theta} = \tan^2\theta = \dfrac{9}{16}\)

यदि \(p\cos A = 2q\sin A\) और \(2p\operatorname{cosec}A – q\sec A = 3\), तो \(p^2 + 4q^2\) का मान है:

\(2\)
\(4\)
\(1\)
\(3\)

Explanation:

\(p\cos A = 2q\sin A \Rightarrow p = 2q\tan A\)
\(2p\operatorname{cosec}A – q\sec A = \dfrac{2p}{\sin A} – \dfrac{q}{\cos A} = \dfrac{4q}{\cos A} – \dfrac{q}{\cos A} = \dfrac{3q}{\cos A} = 3\)
\(q = \cos A\) और \(p = 2q\tan A = 2\sin A\)
\(p^2 + 4q^2 = 4\sin^2 A + 4\cos^2 A = 4\)

यदि \(\dfrac{7\sin\theta + 4\cos\theta}{9\sin\theta – 2\cos\theta} = \dfrac{5}{4}\) है, तो \(\dfrac{\tan^2\theta + 5}{\tan^2\theta – 5}\) का मान ज्ञात कीजिए।

\(2121/769\)
\(-2121/769\)
\(769/2121\)
\(-769/2121\)

Explanation:

अंश व हर को \(\cos\theta\) से भाग देने पर \(\dfrac{7t+4}{9t-2} = \dfrac{5}{4}\), जहाँ \(t = \tan\theta\)
\(28t + 16 = 45t – 10 \Rightarrow 17t = 26 \Rightarrow t = \dfrac{26}{17}\)
\(t^2 = \dfrac{676}{289}\)
\(\dfrac{t^2+5}{t^2-5} = \dfrac{676 + 1445}{676 – 1445} = \dfrac{2121}{-769} = -\dfrac{2121}{769}\)

\(\dfrac{1 + \sin\theta – \cos\theta}{1 + \sin\theta – \cos\theta}\) को हल करें:

\(\cos\dfrac{\theta}{2}\)
\(\sin\dfrac{\theta}{2}\)
\(\cot\dfrac{\theta}{2}\)
\(\tan\dfrac{\theta}{2}\)

Explanation:

प्रश्न में हर में \(+\cos\theta\) होना चाहिए: \(\dfrac{1 + \sin\theta – \cos\theta}{1 + \sin\theta + \cos\theta}\)
\(1 – \cos\theta = 2\sin^2\dfrac{\theta}{2}\), \(\sin\theta = 2\sin\dfrac{\theta}{2}\cos\dfrac{\theta}{2}\), \(1 + \cos\theta = 2\cos^2\dfrac{\theta}{2}\)
अंश \(= 2\sin\dfrac{\theta}{2}\left(\cos\dfrac{\theta}{2} + \sin\dfrac{\theta}{2}\right)\)
हर \(= 2\cos\dfrac{\theta}{2}\left(\sin\dfrac{\theta}{2} + \cos\dfrac{\theta}{2}\right)\)
मान \(= \tan\dfrac{\theta}{2}\)
नोट: प्रश्न में अंश और हर समान छपे हैं (मान \(1\) होता), पर उत्तर कुंजी \(\tan\frac{\theta}{2}\) है, जो हर में \(+\cos\theta\) मानने पर आता है।

\(\left(\dfrac{4\cos(90 – \theta)\sin^3(90 – \theta) – 4\sin(90 + \theta)\cos^3(90 – \theta)}{\cos\dfrac{180 + 8\theta}{2}}\right) = ?\)

\(-1\)
\(1\)
\(2\)
\(0\)

Explanation:

\(\cos(90^\circ – \theta) = \sin\theta,\ \sin(90^\circ+\theta) = \cos\theta,\ \sin(90^\circ – \theta) = \cos\theta\)
अंश \(= 4\sin\theta\cos^3\theta – 4\cos\theta\sin^3\theta = 4\sin\theta\cos\theta(\cos^2\theta – \sin^2\theta)\)
\(= 2\sin 2\theta\cos 2\theta = \sin 4\theta\)
हर \(= \cos\dfrac{180^\circ + 8\theta}{2} = \cos(90^\circ + 4\theta) = -\sin 4\theta\)
मान \(= \dfrac{\sin 4\theta}{-\sin 4\theta} = -1\)

यदि \(\sin 31^\circ = \alpha\) है, तो \(\cot 59^\circ\) का मान ज्ञात कीजिए।

\(\dfrac{\sqrt{1 – \alpha^2}}{2\alpha}\)
\(\dfrac{\alpha}{\sqrt{1 – \alpha^2}}\)
\(\dfrac{\sqrt{1 + \alpha^2}}{\alpha}\)
\(\dfrac{\alpha}{\sqrt{1 + \alpha^2}}\)

Explanation:

\(\cot 59^\circ = \tan 31^\circ = \dfrac{\sin 31^\circ}{\cos 31^\circ}\)
\(\sin 31^\circ = \alpha \Rightarrow \cos 31^\circ = \sqrt{1 – \alpha^2}\)
\(\cot 59^\circ = \dfrac{\alpha}{\sqrt{1 – \alpha^2}}\)

यदि \(3\sin A + 4\cos A = 5\), है, तो \(\tan A\) का मान ज्ञात कीजिए।

\(3/4\)
\(3/5\)
\(5/4\)
\(4/5\)

Explanation:

\(3\sin A + 4\cos A \le \sqrt{3^2 + 4^2} = 5\), अतः मान \(5\) तभी संभव है जब \(\sin A = \dfrac{3}{5}\) और \(\cos A = \dfrac{4}{5}\)
जाँच: \(3\times\dfrac{3}{5} + 4\times\dfrac{4}{5} = \dfrac{9+16}{5} = 5\)
\(\tan A = \dfrac{3}{4}\)

\(\cot^2 A – \dfrac{1}{\sin^2 A} = ?\)

\(0\)
\(-1\)
\(-2\)
\(1\)

Explanation:

\(\cot^2 A – \dfrac{1}{\sin^2 A} = \cot^2 A – \operatorname{cosec}^2 A = -1\)

यदि \((\cos\theta + \sin\theta) : (\cos\theta – \sin\theta) = (\sqrt{3} + 1) : (\sqrt{3} – 1)\), \(0^\circ < \theta < 90^\circ\) है, तो \(\sec\theta\) का मान क्या होगा?

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

Explanation:

\(\dfrac{\cos\theta + \sin\theta}{\cos\theta – \sin\theta} = \dfrac{1 + \tan\theta}{1 – \tan\theta} = \dfrac{\sqrt{3}+1}{\sqrt{3}-1}\)
\((1+t)(\sqrt{3}-1) = (\sqrt{3}+1)(1-t)\)
\(\sqrt{3} – 1 + \sqrt{3}t – t = \sqrt{3} + 1 – \sqrt{3}t – t\)
\(2\sqrt{3}\,t = 2 \Rightarrow t = \dfrac{1}{\sqrt{3}} \Rightarrow \theta = 30^\circ\)
\(\sec 30^\circ = \dfrac{2}{\sqrt{3}} = \dfrac{2\sqrt{3}}{3}\)

निम्न का मान ज्ञात कीजिए। \(\cos(36^\circ – A)\cos(36^\circ + A) + \cos(54^\circ – A)\cos(54^\circ + A)\)

\(\cos A\)
\(\sin 2A\)
\(\cos 2A\)
\(\sin A\)

Explanation:

\(\cos(36^\circ – A)\cos(36^\circ + A) = \dfrac{1}{2}\left[\cos 72^\circ + \cos 2A\right]\)
\(\cos(54^\circ – A)\cos(54^\circ + A) = \dfrac{1}{2}\left[\cos 108^\circ + \cos 2A\right]\)
जोड़ने पर \(\cos 72^\circ + \cos 108^\circ = 0\), अतः मान \(= \cos 2A\)

यदि \(\cot\theta = \dfrac{4}{3}\), \(0 < \theta < \dfrac{\pi}{2}\) और \(5p\cos^2\theta\sin\theta = \cot^2\theta\) है, तो \(p\) का मान है:

\(\dfrac{7}{27}\)
\(\dfrac{125}{27}\)
\(\dfrac{5}{27}\)
\(\dfrac{25}{27}\)

Explanation:

\(\cot\theta = \dfrac{4}{3} \Rightarrow \sin\theta = \dfrac{3}{5},\ \cos\theta = \dfrac{4}{5}\)
\(5p\cos^2\theta\sin\theta = 5p\times\dfrac{16}{25}\times\dfrac{3}{5} = \dfrac{48p}{25}\)
\(\cot^2\theta = \dfrac{16}{9}\)
\(\dfrac{48p}{25} = \dfrac{16}{9} \Rightarrow p = \dfrac{16\times 25}{9\times 48} = \dfrac{25}{27}\)

\(\Delta ABC\) में, \(\angle B = 90^\circ\) और \(AB : BC = 1 : 2\). \(\cos A + \tan C\) का मान है:

\(\dfrac{5 + \sqrt{5}}{2\sqrt{5}}\)
\(\dfrac{1 + \sqrt{5}}{2\sqrt{5}}\)
\(\dfrac{2\sqrt{5}}{2 + \sqrt{5}}\)
\(\dfrac{2 + \sqrt{5}}{2\sqrt{5}}\)

Explanation:

माना \(AB = k,\ BC = 2k\); तब \(AC = \sqrt{k^2 + 4k^2} = \sqrt{5}k\)
\(\cos A = \dfrac{AB}{AC} = \dfrac{1}{\sqrt{5}}\)
\(\tan C = \dfrac{AB}{BC} = \dfrac{1}{2}\)
\(\cos A + \tan C = \dfrac{1}{\sqrt{5}} + \dfrac{1}{2} = \dfrac{2+\sqrt{5}}{2\sqrt{5}}\)

यदि \(\cos A = \dfrac{1}{2}\), \(0^\circ \le A \le 90^\circ\) है, तो \(\sin(180 – A)\) का मान क्या है?

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

Explanation:

\(\cos A = \dfrac{1}{2} \Rightarrow A = 60^\circ\)
\(\sin(180^\circ – A) = \sin A = \sin 60^\circ = \dfrac{\sqrt{3}}{2}\)

यदि \(\sin\beta = \dfrac{1}{3}\), तो \((\sec\beta – \tan\beta)^2\) इसके बराबर है:

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

Explanation:

\((\sec\beta – \tan\beta)^2 = \dfrac{(1-\sin\beta)^2}{\cos^2\beta} = \dfrac{(1-\sin\beta)^2}{(1-\sin\beta)(1+\sin\beta)} = \dfrac{1-\sin\beta}{1+\sin\beta}\)
\(= \dfrac{1 – \frac{1}{3}}{1 + \frac{1}{3}} = \dfrac{2/3}{4/3} = \dfrac{1}{2}\)

यदि \(\sin A = \dfrac{\sqrt{3}}{2}\), \(0^\circ < A < 90^\circ\) है, तो \(2(\operatorname{cosec}A + \cot A)\) का मान ज्ञात कीजिए।

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

Explanation:

\(\sin A = \dfrac{\sqrt{3}}{2} \Rightarrow A = 60^\circ\)
\(\operatorname{cosec}60^\circ = \dfrac{2}{\sqrt{3}},\ \cot 60^\circ = \dfrac{1}{\sqrt{3}}\)
\(2(\operatorname{cosec}A + \cot A) = 2\times\dfrac{3}{\sqrt{3}} = 2\sqrt{3}\)

यदि \(\sin A = \dfrac{2}{3}\) है, तो \((7 – \tan A)(3 + \cos A)\) का मान ज्ञात कीजिए।

\(\dfrac{61}{3} + \dfrac{17}{3\sqrt{5}}\)
\(\dfrac{61}{3\sqrt{5}} + \dfrac{17}{3}\)
\(\dfrac{61}{3} + \dfrac{17}{\sqrt{5}}\)
\(\dfrac{61}{3} – \dfrac{17}{3\sqrt{5}}\)

Explanation:

\(\sin A = \dfrac{2}{3} \Rightarrow \cos A = \dfrac{\sqrt{5}}{3},\ \tan A = \dfrac{2}{\sqrt{5}}\)
\((7 – \tan A)(3 + \cos A) = 21 + \dfrac{7\sqrt{5}}{3} – \dfrac{6}{\sqrt{5}} – \dfrac{2}{3}\)
\(\dfrac{7\sqrt{5}}{3} – \dfrac{6\sqrt{5}}{5} = \dfrac{35\sqrt{5} – 18\sqrt{5}}{15} = \dfrac{17\sqrt{5}}{15} = \dfrac{17}{3\sqrt{5}}\)
मान \(= \dfrac{61}{3} + \dfrac{17}{3\sqrt{5}}\)

यदि \(\dfrac{2\sin A – \cos A}{\sin A + \cos A} = 1\) तो \(\cot A\) का मान ज्ञात कीजिए।

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

Explanation:

\(2\sin A – \cos A = \sin A + \cos A \Rightarrow \sin A = 2\cos A\)
\(\tan A = 2 \Rightarrow \cot A = \dfrac{1}{2}\)

यदि \(\sin\alpha = \dfrac{5}{13}\) है, तो \(\cos\alpha \cdot \operatorname{cosec}\alpha \cdot \cot\alpha\) का मान ________ है।

\(\dfrac{12}{5}\)
\(\dfrac{144}{25}\)
\(\dfrac{5}{12}\)
\(\dfrac{25}{144}\)

Explanation:

\(\sin\alpha = \dfrac{5}{13} \Rightarrow \cos\alpha = \dfrac{12}{13}\)
\(\cos\alpha\cdot\operatorname{cosec}\alpha\cdot\cot\alpha = \dfrac{\cos\alpha}{\sin\alpha}\cdot\cot\alpha = \cot^2\alpha\)
\(\cot^2\alpha = \left(\dfrac{12}{5}\right)^2 = \dfrac{144}{25}\)

यदि \(\dfrac{\sec A + \tan A}{\sec A – \tan A} = 2\dfrac{51}{79}\), तो \(\sin A\) का मान बराबर है:

\(\dfrac{77}{144}\)
\(\dfrac{61}{169}\)
\(\dfrac{87}{169}\)
\(\dfrac{65}{144}\)

Explanation:

\(\dfrac{\sec A + \tan A}{\sec A – \tan A} = \dfrac{1 + \sin A}{1 – \sin A}\) (अंश व हर को \(\cos A\) से गुणा करने पर)
\(2\dfrac{51}{79} = \dfrac{209}{79}\)
\(79(1+\sin A) = 209(1 – \sin A) \Rightarrow 288\sin A = 130\)
\(\sin A = \dfrac{65}{144}\)

यदि \(\cot\theta = \sqrt{6}\) है, तो \(\dfrac{\operatorname{cosec}^2\theta + \sec^2\theta}{\operatorname{cosec}^2\theta – \sec^2\theta}\) का मान होगा:

\(\dfrac{49}{36}\)
\(\dfrac{43}{36}\)
\(\dfrac{7}{5}\)
\(\dfrac{48}{35}\)

Explanation:

\(\cot\theta = \sqrt{6} \Rightarrow \operatorname{cosec}^2\theta = 1 + 6 = 7\)
\(\tan^2\theta = \dfrac{1}{6} \Rightarrow \sec^2\theta = \dfrac{7}{6}\)
\(\dfrac{7 + \frac{7}{6}}{7 – \frac{7}{6}} = \dfrac{49/6}{35/6} = \dfrac{7}{5}\)

यदि \(\cos x = -\dfrac{\sqrt{3}}{2}\) और \(\pi < x < \dfrac{3\pi}{2}\) है, तो \(2\cot^2 x + 3\sec^2 x\) का मान है?

\(10\)
\(4\)
\(8\)
\(16\)

Explanation:

\(\cos x = -\dfrac{\sqrt{3}}{2}\) और \(x\) तीसरे चतुर्थांश में है, अतः \(\sin x = -\dfrac{1}{2}\)
\(\cot^2 x = \dfrac{3/4}{1/4} = 3,\ \sec^2 x = \dfrac{4}{3}\)
\(2\cot^2 x + 3\sec^2 x = 6 + 4 = 10\)

निम्नलिखित में से कौन \(0^\circ < \theta < 90^\circ\) की रेंज में निरंतर बढ़ता है?

\(\cot\theta\)
\(\operatorname{cosec}\theta\)
\(\tan\theta\)
\(\cos\theta\)

Explanation:

\(0^\circ \lt \theta \lt 90^\circ\) में \(\tan\theta\) का मान \(0\) से \(\infty\) तक निरंतर बढ़ता है।
\(\cot\theta,\ \cos\theta\) घटते हैं और \(\operatorname{cosec}\theta\) भी घटता है (\(\infty\) से \(1\) तक)।
अतः उत्तर \(\tan\theta\)।

\(\dfrac{\sin^3 21^\circ + \cos^3 19^\circ}{\sin 21^\circ + \cos 19^\circ} + \sin^2 69^\circ + \cos^2 71^\circ + \dfrac{1}{\sec 69^\circ \operatorname{cosec}71^\circ} = ?\)

\(3\)
\(1\)
\(4\)
\(2\)

Explanation:

\(\sin^3 21^\circ + \cos^3 19^\circ\) को \(a^3+b^3\) मानें (\(a = \sin 21^\circ,\ b = \cos 19^\circ\)):
\(\dfrac{a^3+b^3}{a+b} = a^2 – ab + b^2 = \sin^2 21^\circ – \sin 21^\circ\cos 19^\circ + \cos^2 19^\circ\)
\(\sin 69^\circ = \cos 21^\circ,\ \cos 71^\circ = \sin 19^\circ\)
\(\dfrac{1}{\sec 69^\circ\operatorname{cosec}71^\circ} = \cos 69^\circ\sin 71^\circ = \sin 21^\circ\cos 19^\circ\)
कुल \(= \sin^2 21^\circ + \cos^2 19^\circ + \cos^2 21^\circ + \sin^2 19^\circ = 1 + 1 = 2\)

यदि \(\theta = 9^\circ\) तो \(\cot\theta \cot 2\theta \cot 3\theta \cot 4\theta \cot 5\theta \cot 6\theta \cot 7\theta \cot 8\theta \cot 9\theta\) का मान क्या होगा?

\(\sqrt{3} – 1\)
\(1\)
\(\sqrt{3}\)
\(\dfrac{1}{\sqrt{3}}\)

Explanation:

\(\theta = 9^\circ\) पर पद हैं \(\cot 9^\circ,\cot 18^\circ,\dots,\cot 81^\circ\)
\(\cot x\cdot\cot(90^\circ – x) = 1\) के अनुसार \(\cot 9^\circ\cot 81^\circ = \cot 18^\circ\cot 72^\circ = \cot 27^\circ\cot 63^\circ = \cot 36^\circ\cot 54^\circ = 1\)
बीच का पद \(\cot 45^\circ = 1\)
गुणनफल \(= 1\)

यदि \(\cos x = \dfrac{3}{5}\) है, तो \(\sin x – \sin^3 x\) का मान ज्ञात करें।

0.476
0.389
0.358
0.288

Explanation:

\(\cos x = \dfrac{3}{5} \Rightarrow \sin x = \dfrac{4}{5}\)
\(\sin x – \sin^3 x = \sin x(1 – \sin^2 x) = \sin x\cos^2 x\)
\(= \dfrac{4}{5}\times\dfrac{9}{25} = \dfrac{36}{125} = 0.288\)

यदि \(\sin\theta = 0.96\) तो \(\cot\theta\) का मान ज्ञात करें।

0.4563
0.4876
0.2916
0.3456

Explanation:

\(\sin\theta = 0.96 = \dfrac{24}{25} \Rightarrow \cos\theta = \dfrac{7}{25}\)
\(\cot\theta = \dfrac{7}{24} = 0.29166\ldots \approx 0.2916\)

यदि \(2\cot\theta = 3\) है, तो \(\dfrac{\sqrt{13}\cos\theta – 3\tan\theta}{3\tan\theta + \sqrt{13}\sin\theta}\) का मान क्या होगा?

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

Explanation:

\(2\cot\theta = 3 \Rightarrow \cot\theta = \dfrac{3}{2} \Rightarrow \sin\theta = \dfrac{2}{\sqrt{13}},\ \cos\theta = \dfrac{3}{\sqrt{13}},\ \tan\theta = \dfrac{2}{3}\)
अंश \(= \sqrt{13}\times\dfrac{3}{\sqrt{13}} – 3\times\dfrac{2}{3} = 3 – 2 = 1\)
हर \(= 3\times\dfrac{2}{3} + \sqrt{13}\times\dfrac{2}{\sqrt{13}} = 2 + 2 = 4\)
मान \(= \dfrac{1}{4}\)

\(\tan(63^\circ – \theta) – \cot(27^\circ + \theta) + \dfrac{\operatorname{cosec}^2 70^\circ – \tan^2 20^\circ}{\sec^2 37^\circ – \cot^2 53^\circ} = ?\)

\(2\)
\(3\)
\(0\)
\(1\)

Explanation:

\((63^\circ – \theta) + (27^\circ + \theta) = 90^\circ\), अतः \(\tan(63^\circ – \theta) = \cot(27^\circ+\theta)\) और इनका अंतर \(0\) है।
\(\operatorname{cosec}^2 70^\circ – \tan^2 20^\circ = \sec^2 20^\circ – \tan^2 20^\circ = 1\)
\(\sec^2 37^\circ – \cot^2 53^\circ = \sec^2 37^\circ – \tan^2 37^\circ = 1\)
मान \(= 0 + \dfrac{1}{1} = 1\)

यदि \(\sqrt{3}\cos\theta = \sin\theta\) है, तो \(\dfrac{4\sin^2\theta – 5\cos\theta}{3\cos\theta + 1}\) का मान ज्ञात करें।

\(\dfrac{1}{4}\)
\(\dfrac{1}{5}\)
\(5\)
\(\dfrac{2}{5}\)

Explanation:

\(\sqrt{3}\cos\theta = \sin\theta \Rightarrow \tan\theta = \sqrt{3} \Rightarrow \theta = 60^\circ\)
\(\sin\theta = \dfrac{\sqrt{3}}{2},\ \cos\theta = \dfrac{1}{2}\)
\(\dfrac{4\times\frac{3}{4} – \frac{5}{2}}{\frac{3}{2} + 1} = \dfrac{1/2}{5/2} = \dfrac{1}{5}\)

यदि \(\cot\theta = \dfrac{3}{\sqrt{5}}\), \(0^\circ < \theta < 90^\circ\) है, तो \(\dfrac{6\sec^2\theta – \frac{5}{3}\operatorname{cosec}^2\theta}{\frac{3}{5}\sec^2\theta + \frac{4}{3}\operatorname{cosec}^2\theta}\) का मान है:

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

Explanation:

\(\cot\theta = \dfrac{3}{\sqrt{5}} \Rightarrow \operatorname{cosec}^2\theta = 1 + \dfrac{9}{5} = \dfrac{14}{5}\)
\(\tan^2\theta = \dfrac{5}{9} \Rightarrow \sec^2\theta = \dfrac{14}{9}\)
अंश \(= 6\times\dfrac{14}{9} – \dfrac{5}{3}\times\dfrac{14}{5} = \dfrac{28}{3} – \dfrac{14}{3} = \dfrac{14}{3}\)
हर \(= \dfrac{3}{5}\times\dfrac{14}{9} + \dfrac{4}{3}\times\dfrac{14}{5} = \dfrac{14}{15} + \dfrac{56}{15} = \dfrac{14}{3}\)
मान \(= 1\)

यदि \(\cos 27^\circ = x\), तो \(\tan 63^\circ\) का मान है:

\(\dfrac{\sqrt{1 + x^2}}{x}\)
\(\dfrac{x}{\sqrt{1 + x^2}}\)
\(\dfrac{\sqrt{1 – x^2}}{x}\)
\(\dfrac{x}{\sqrt{1 – x^2}}\)

Explanation:

\(\tan 63^\circ = \cot 27^\circ = \dfrac{\cos 27^\circ}{\sin 27^\circ}\)
\(\cos 27^\circ = x \Rightarrow \sin 27^\circ = \sqrt{1 – x^2}\)
\(\tan 63^\circ = \dfrac{x}{\sqrt{1 – x^2}}\)

यदि \(3\cot A = 4\tan A\), तथा A एक न्यून कोण है, तो \(\sec A\) का मान ज्ञात करें।

\(\dfrac{\sqrt{7}}{2}\)
\(\dfrac{\sqrt{21}}{3}\)
\(\dfrac{1}{\sqrt{3}}\)
\(\dfrac{1}{2}\)

Explanation:

\(3\cot A = 4\tan A \Rightarrow 3 = 4\tan^2 A \Rightarrow \tan^2 A = \dfrac{3}{4}\)
\(\sec^2 A = 1 + \tan^2 A = \dfrac{7}{4}\)
\(\sec A = \dfrac{\sqrt{7}}{2}\) (न्यून कोण)

यदि \(\operatorname{cosec}\theta = \dfrac{\sqrt{5}}{2}\), तो \((\sec\theta + \tan\theta – \cot\theta \sin\theta)\) का मान ज्ञात करें।

\(2 + \dfrac{4\sqrt{5}}{5}\)
\(2 + \sqrt{5}\)
\(2 + \dfrac{\sqrt{5}}{2}\)
\(2 + \dfrac{2\sqrt{5}}{5}\)

Explanation:

\(\operatorname{cosec}\theta = \dfrac{\sqrt{5}}{2} \Rightarrow \sin\theta = \dfrac{2}{\sqrt{5}},\ \cos\theta = \dfrac{1}{\sqrt{5}}\)
\(\sec\theta = \sqrt{5},\ \tan\theta = 2,\ \cot\theta = \dfrac{1}{2}\)
\(\sec\theta + \tan\theta – \cot\theta\sin\theta = \sqrt{5} + 2 – \dfrac{1}{2}\times\dfrac{2}{\sqrt{5}}\)
\(= 2 + \sqrt{5} – \dfrac{1}{\sqrt{5}} = 2 + \dfrac{5 – 1}{\sqrt{5}} = 2 + \dfrac{4\sqrt{5}}{5}\)

यदि \(\sec\theta = \dfrac{65}{63}\) और \(\theta\) न्यून कोण है, तो \(8(\operatorname{cosec}\theta – \cot\theta)\) का मान ज्ञात करें।

\(8\)
\(4\)
\(2\)
\(1\)

Explanation:

\(\sec\theta = \dfrac{65}{63} \Rightarrow \cos\theta = \dfrac{63}{65},\ \sin\theta = \dfrac{16}{65}\)
\(\operatorname{cosec}\theta – \cot\theta = \dfrac{1 – \cos\theta}{\sin\theta} = \dfrac{2/65}{16/65} = \dfrac{1}{8}\)
\(8\times\dfrac{1}{8} = 1\)

यदि \(\operatorname{cosec}A = 10\) है, तो \(20\sin A + 9\sqrt{11}\sec A\) का मान ज्ञात करें। दिया गया है कि, A न्यून कोण है।

\(23\)
\(34\)
\(32\)
\(30\)

Explanation:

\(\operatorname{cosec}A = 10 \Rightarrow \sin A = \dfrac{1}{10},\ \cos A = \dfrac{\sqrt{99}}{10} = \dfrac{3\sqrt{11}}{10}\)
\(\sec A = \dfrac{10}{3\sqrt{11}}\)
\(20\sin A + 9\sqrt{11}\sec A = 2 + 9\sqrt{11}\times\dfrac{10}{3\sqrt{11}} = 2 + 30 = 32\)

यदि त्रिभुज ABC, A पर समकोण है, तो \(\sin\dfrac{B + C}{2}\cos\dfrac{B + C}{2}\) का मान ज्ञात करें।

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

Explanation:

\(A = 90^\circ \Rightarrow B + C = 90^\circ \Rightarrow \dfrac{B+C}{2} = 45^\circ\)
\(\sin 45^\circ\cos 45^\circ = \dfrac{1}{\sqrt{2}}\times\dfrac{1}{\sqrt{2}} = \dfrac{1}{2}\)

यदि \(\cot\theta = \sqrt{2} + 1\) है, तो \(\operatorname{cosec}\theta \sec\theta\) का मान ज्ञात करें।

\(\dfrac{\sqrt{2}}{2}\)
\(4\sqrt{2}\)
\(\dfrac{\sqrt{2}}{4}\)
\(2\sqrt{2}\)

Explanation:

\(\operatorname{cosec}\theta\sec\theta = \dfrac{1}{\sin\theta\cos\theta} = \tan\theta + \cot\theta\)
\(\cot\theta = \sqrt{2}+1 \Rightarrow \tan\theta = \dfrac{1}{\sqrt{2}+1} = \sqrt{2}-1\)
\(\tan\theta + \cot\theta = (\sqrt{2}-1) + (\sqrt{2}+1) = 2\sqrt{2}\)

यदि \(\tan A = \dfrac{1.1}{6}\) है, तो \((4\cos A – 7\sin A)\) का मान क्या होगा? दिया गया है कि A एक न्यून कोण है।

\(2\dfrac{14}{61}\)
\(2\dfrac{41}{71}\)
\(2\dfrac{14}{71}\)
\(2\dfrac{41}{61}\)

Explanation:

\(\tan A = \dfrac{1.1}{6} = \dfrac{11}{60}\)
कर्ण \(= \sqrt{11^2 + 60^2} = \sqrt{3721} = 61\)
\(\sin A = \dfrac{11}{61},\ \cos A = \dfrac{60}{61}\)
\(4\cos A – 7\sin A = \dfrac{240 – 77}{61} = \dfrac{163}{61} = 2\dfrac{41}{61}\)

यदि A का मान \(45^\circ\) और \(540^\circ\) के मध्य है, और \(\sin A = 0.5\) है, तो \(A/3\) का मान अंश में कितना होगा?

\(170^\circ\)
\(175^\circ\)
\(165^\circ\)
\(160^\circ\)

Explanation:

\(\sin A = 0.5\) के हल: \(A = 30^\circ,\ 150^\circ,\ 390^\circ,\ 510^\circ,\dots\)
\(45^\circ \lt A \lt 540^\circ\) में संभव मान हैं \(150^\circ,\ 390^\circ,\ 510^\circ\)
\(\dfrac{A}{3}\) के मान \(50^\circ,\ 130^\circ,\ 170^\circ\); विकल्पों में केवल \(170^\circ\) है।
\(A = 510^\circ\) और \(\dfrac{A}{3} = 170^\circ\)

यदि \(y = \tan 35^\circ\) है, तो \((2\tan 55^\circ + \cot 55^\circ)\) का मान बताएँ।

\(\dfrac{2 + y^2}{y}\)
\(\dfrac{2 – y}{y^3}\)
\(\dfrac{2}{y^2}\)
\(\dfrac{2 – y^2}{y}\)

Explanation:

\(\tan 55^\circ = \cot 35^\circ = \dfrac{1}{\tan 35^\circ} = \dfrac{1}{y}\) और \(\cot 55^\circ = \tan 35^\circ = y\)
\(2\tan 55^\circ + \cot 55^\circ = \dfrac{2}{y} + y = \dfrac{2 + y^2}{y}\)

यदि \(\sin\theta = \dfrac{2\sqrt{ab}}{a + b}\) है, \(a > b > 0\) है, तो \(\dfrac{\cos\theta + 1}{\cos\theta – 1}\) का मान ज्ञात करें।

\(-\dfrac{b}{a}\)
\(-\dfrac{a}{b}\)
\(\dfrac{a}{b}\)
\(\dfrac{b}{a}\)

Explanation:

\(\sin\theta = \dfrac{2\sqrt{ab}}{a+b} \Rightarrow \cos\theta = \sqrt{1 – \dfrac{4ab}{(a+b)^2}} = \dfrac{a-b}{a+b}\) (\(a \gt b\))
\(\cos\theta + 1 = \dfrac{2a}{a+b},\ \cos\theta – 1 = -\dfrac{2b}{a+b}\)
मान \(= -\dfrac{a}{b}\)

एक \(\Delta ABC\) में, B पर समकोण है, यदि \(\tan A = \dfrac{1}{\sqrt{3}}\) है, तो \(\sin A\cos C + \cos A \sin C\) का मान ज्ञात कीजिए।

\(0\)
\(2\)
\(-1\)
\(1\)

Explanation:

\(\tan A = \dfrac{1}{\sqrt{3}} \Rightarrow A = 30^\circ\); \(B = 90^\circ\) अतः \(C = 60^\circ\)
\(\sin A\cos C + \cos A\sin C = \sin(A + C) = \sin 90^\circ = 1\)

यदि \(\alpha\) एक न्यून कोण है, तो निम्नलिखित में से कौन सा विकल्प आवश्यक रूप से \(\sec\alpha\) के मान के बराबर नहीं होगा।

\(\dfrac{1}{\cos\alpha}\)
\(\dfrac{1}{\sin\alpha}\)
\(\sqrt{1 + \tan^2\alpha}\)
\(\dfrac{\tan\alpha}{\sin\alpha}\)

Explanation:

\(\dfrac{1}{\cos\alpha} = \sec\alpha\), \(\sqrt{1+\tan^2\alpha} = \sec\alpha\), \(\dfrac{\tan\alpha}{\sin\alpha} = \dfrac{1}{\cos\alpha} = \sec\alpha\)
ये तीनों सदैव \(\sec\alpha\) के बराबर हैं।
\(\dfrac{1}{\sin\alpha} = \operatorname{cosec}\alpha\), जो \(\sec\alpha\) के बराबर आवश्यक नहीं है।

यदि \(\alpha\) एक न्यून कोण है, तो निम्नलिखित में से कौन सा विकल्प अनिवार्यतः \(\operatorname{cosec}\alpha\) के मान के बराबर नहीं होगा?

\(\dfrac{1}{\cos\alpha}\)
\(\sqrt{1 + \cot^2\alpha}\)
\(\dfrac{1}{\sin\alpha}\)
\(\dfrac{\cot\alpha}{\cos\alpha}\)

Explanation:

\(\dfrac{1}{\sin\alpha}\), \(\sqrt{1+\cot^2\alpha}\) और \(\dfrac{\cot\alpha}{\cos\alpha} = \dfrac{1}{\sin\alpha}\) सदैव \(\operatorname{cosec}\alpha\) के बराबर हैं।
\(\dfrac{1}{\cos\alpha} = \sec\alpha\), जो \(\operatorname{cosec}\alpha\) के बराबर आवश्यक नहीं है।

यदि \(\tan A + \sec A = K\), तो \(\sin A\) का मान क्या है?

\(\dfrac{2K}{K^2 + 1}\)
\(\dfrac{K^2 – 1}{2K^2}\)
\(\dfrac{K^2 + 1}{K^2 – 1}\)
\(\dfrac{K^2 – 1}{K^2 + 1}\)

Explanation:

\((\sec A + \tan A)(\sec A – \tan A) = 1 \Rightarrow \sec A – \tan A = \dfrac{1}{K}\)
\(\sec A = \dfrac{1}{2}\left(K + \dfrac{1}{K}\right) = \dfrac{K^2+1}{2K}\)
\(\tan A = \dfrac{1}{2}\left(K – \dfrac{1}{K}\right) = \dfrac{K^2-1}{2K}\)
\(\sin A = \dfrac{\tan A}{\sec A} = \dfrac{K^2 – 1}{K^2 + 1}\)

यदि \(\operatorname{cosec}\theta + \cot\theta = s\) है, तो \(\cos\theta\) का मान क्या है?

\(\dfrac{s^2 + 1}{s^2 – 1}\)
\(\dfrac{s^2 – 1}{2s}\)
\(\dfrac{2s}{s^2 + 1}\)
\(\dfrac{s^2 – 1}{s^2 + 1}\)

Explanation:

\((\operatorname{cosec}\theta + \cot\theta)(\operatorname{cosec}\theta – \cot\theta) = 1 \Rightarrow \operatorname{cosec}\theta – \cot\theta = \dfrac{1}{s}\)
\(\operatorname{cosec}\theta = \dfrac{s^2+1}{2s},\ \cot\theta = \dfrac{s^2-1}{2s}\)
\(\cos\theta = \dfrac{\cot\theta}{\operatorname{cosec}\theta} = \dfrac{s^2 – 1}{s^2 + 1}\)

यदि \(\sec\theta – \tan\theta = m\), तो \(2\tan\theta = ?\)

\(\dfrac{1 – m}{m}\)
\(\dfrac{1 – m^2}{m}\)
\(\dfrac{1 + m^2}{m}\)
\(\dfrac{m^2 – 1}{m}\)

Explanation:

\(\sec\theta – \tan\theta = m \Rightarrow \sec\theta + \tan\theta = \dfrac{1}{m}\)
घटाने पर \(2\tan\theta = \dfrac{1}{m} – m = \dfrac{1 – m^2}{m}\)

यदि \(\sec\theta = 2\dfrac{4}{23}\) है, तो \(\tan^2\theta\) का मान ज्ञात करें?

\(3\dfrac{177}{529}\)
\(2\dfrac{16}{529}\)
\(1\dfrac{200}{529}\)
\(3\dfrac{384}{529}\)

Explanation:

\(\sec\theta = 2\dfrac{4}{23} = \dfrac{50}{23}\)
\(\tan^2\theta = \sec^2\theta – 1 = \dfrac{2500}{529} – 1 = \dfrac{1971}{529} = 3\dfrac{384}{529}\)

यदि \(\theta\) है, तो \(\cos\theta = \dfrac{39}{89}\) का मान ज्ञात करें?

\(\dfrac{49}{89}\)
\(\dfrac{79}{89}\)
\(\dfrac{80}{89}\)
\(\dfrac{50}{89}\)

Explanation:

प्रश्न में \(\sin\theta\) का मान पूछा गया प्रतीत होता है (प्रश्न का पाठ अधूरा है): \(\cos\theta = \dfrac{39}{89}\) दिया है।
\(\sin\theta = \sqrt{1 – \dfrac{39^2}{89^2}} = \dfrac{\sqrt{89^2 – 39^2}}{89} = \dfrac{\sqrt{7921 – 1521}}{89} = \dfrac{\sqrt{6400}}{89} = \dfrac{80}{89}\)
नोट: प्रश्न में “\(\sin\theta\) का मान” छूटा हुआ है; उसी के अनुसार हल किया गया है, जो कुंजी से मेल खाता है।

यदि \(\theta\) एक न्यूनकोण है और \(\sin\theta = \dfrac{43}{47}\) है, तो \(\cos\theta\) का मान ज्ञात करें?

\(\dfrac{43}{6\sqrt{10}}\)
\(\dfrac{47}{6\sqrt{10}}\)
\(\dfrac{6\sqrt{10}}{43}\)
\(\dfrac{6\sqrt{10}}{47}\)

Explanation:

\(\sin\theta = \dfrac{43}{47} \Rightarrow \cos\theta = \dfrac{\sqrt{47^2 – 43^2}}{47}\)
\(47^2 – 43^2 = 2209 – 1849 = 360\)
\(\cos\theta = \dfrac{\sqrt{360}}{47} = \dfrac{6\sqrt{10}}{47}\)

यदि \(\cot A = \dfrac{5}{12}\) है, तो \(\sqrt{\dfrac{\sec A – \operatorname{cosec}A}{\sec A + \operatorname{cosec}A}} – \sqrt{\dfrac{17}{7}}\) का मान ज्ञात कीजिए।

\(\dfrac{25}{\sqrt{119}}\)
\(0\)
\(-\dfrac{10}{\sqrt{119}}\)
\(\dfrac{10}{\sqrt{119}}\)

Explanation:

\(\cot A = \dfrac{5}{12} \Rightarrow \sin A = \dfrac{12}{13},\ \cos A = \dfrac{5}{13}\)
\(\sec A = \dfrac{13}{5},\ \operatorname{cosec}A = \dfrac{13}{12}\)
\(\dfrac{\sec A – \operatorname{cosec}A}{\sec A + \operatorname{cosec}A} = \dfrac{\frac{1}{5} – \frac{1}{12}}{\frac{1}{5} + \frac{1}{12}} = \dfrac{7/60}{17/60} = \dfrac{7}{17}\)
\(\sqrt{\dfrac{7}{17}} – \sqrt{\dfrac{17}{7}} = \dfrac{7}{\sqrt{119}} – \dfrac{17}{\sqrt{119}} = -\dfrac{10}{\sqrt{119}}\)

यदि \(\sqrt{13}\sin\theta = 2\), तो \(\dfrac{3\tan\theta + \sqrt{13}\sin\theta}{\sqrt{13}\cos\theta – 3\tan\theta}\) ज्ञात कीजिए?

\(4\)
\(3\)
\(2\)
\(0\)

Explanation:

\(\sqrt{13}\sin\theta = 2 \Rightarrow \sin\theta = \dfrac{2}{\sqrt{13}},\ \cos\theta = \dfrac{3}{\sqrt{13}},\ \tan\theta = \dfrac{2}{3}\)
अंश \(= 3\times\dfrac{2}{3} + 2 = 4\)
हर \(= \sqrt{13}\times\dfrac{3}{\sqrt{13}} – 2 = 1\)
मान \(= 4\)

समकोण त्रिभुज ABC में, \(\angle B = 90^\circ\) तथा कोण A और कोण C न्यून कोण हैं। यदि \(\operatorname{cosec}A = 2\sqrt{2}\) है, तो \(\sin A \cdot \cos C + \cos A \cdot \sin C\) ज्ञात कीजिए।

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

Explanation:

\(\angle B = 90^\circ \Rightarrow A + C = 90^\circ\)
\(\sin A\cos C + \cos A\sin C = \sin(A + C) = \sin 90^\circ = 1\)

दिया गया है \(\sin\theta = \dfrac{15}{17}\)। \(\dfrac{15\cot\theta + 17\sin\theta}{8\tan\theta + 16\sec\theta}\) का मूल्यांकन करें, जहाँ \(0^\circ < \theta < 90^\circ\)

\(\dfrac{23}{49}\)
\(\dfrac{49}{23}\)
\(\dfrac{49}{26}\)
\(\dfrac{26}{23}\)

Explanation:

\(\sin\theta = \dfrac{15}{17} \Rightarrow \cos\theta = \dfrac{8}{17},\ \cot\theta = \dfrac{8}{15},\ \tan\theta = \dfrac{15}{8},\ \sec\theta = \dfrac{17}{8}\)
अंश \(= 15\times\dfrac{8}{15} + 17\times\dfrac{15}{17} = 8 + 15 = 23\)
हर \(= 8\times\dfrac{15}{8} + 16\times\dfrac{17}{8} = 15 + 34 = 49\)
मान \(= \dfrac{23}{49}\)

निम्नलिखित व्यंजक \(\sin(30 + \theta) – \cos(60 – \theta)\) का मान ज्ञात कीजिए।

\(2\sin(30 + \theta)\)
\(1\)
\(\dfrac{\sqrt{3}}{2}\)
\(0\)

Explanation:

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

यदि \(\tan\theta = \dfrac{7}{4}\) है, तो \(\dfrac{4\sin\theta – 3\cos\theta}{4\sin\theta + 3\cos\theta}\) का मान ज्ञात करें।

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

Explanation:

अंश व हर को \(\cos\theta\) से भाग देने पर
\(\dfrac{4\tan\theta – 3}{4\tan\theta + 3} = \dfrac{7 – 3}{7 + 3} = \dfrac{4}{10} = \dfrac{2}{5}\)

यदि \(\sin\theta = \dfrac{2xy}{x^2 + y^2}\) है, तो \(\tan\theta\) का मान ज्ञात कीजिये।

\(\dfrac{x^2 + y^2}{x^2 – y^2}\)
\(\dfrac{2xy}{x^2 – y^2}\)
\(\dfrac{x^2 – y^2}{2xy}\)
\(\dfrac{x^2 – y^2}{x^2 + y^2}\)

Explanation:

\(\sin\theta = \dfrac{2xy}{x^2+y^2} \Rightarrow \cos\theta = \dfrac{x^2 – y^2}{x^2 + y^2}\) (\(x \gt y\))
\(\tan\theta = \dfrac{\sin\theta}{\cos\theta} = \dfrac{2xy}{x^2 – y^2}\)

यदि \(\sin\theta = \dfrac{11}{12}\) है, तो \(\tan\theta + \sec\theta\) का मान ज्ञात कीजिए।

\(\sqrt{23}\)
\(\dfrac{11}{\sqrt{23}}\)
\(\dfrac{12}{\sqrt{23}}\)
\(23\)

Explanation:

\(\sin\theta = \dfrac{11}{12} \Rightarrow \cos\theta = \dfrac{\sqrt{144 – 121}}{12} = \dfrac{\sqrt{23}}{12}\)
\(\tan\theta + \sec\theta = \dfrac{11}{\sqrt{23}} + \dfrac{12}{\sqrt{23}} = \dfrac{23}{\sqrt{23}} = \sqrt{23}\)

यदि \(\theta\) न्यून कोण हो और \(\cos\theta = \dfrac{24}{25}\) हो, तो \(\cot(90 – \theta)\) का मान ज्ञात कीजिए।

\(\dfrac{5}{24}\)
\(\dfrac{7}{24}\)
\(\dfrac{7}{25}\)
\(\dfrac{6}{25}\)

Explanation:

\(\cot(90^\circ – \theta) = \tan\theta\)
\(\cos\theta = \dfrac{24}{25} \Rightarrow \sin\theta = \dfrac{7}{25}\)
\(\tan\theta = \dfrac{7}{24}\)

यदि \(t\) एक न्यून कोण है और \(\tan(t) = \dfrac{3}{7}\) है, तो \(\operatorname{cosec}(t)\) का मान क्या है?

\(\dfrac{\sqrt{53}}{7}\)
\(\dfrac{\sqrt{53}}{3}\)
\(\dfrac{7}{\sqrt{58}}\)
\(\dfrac{\sqrt{58}}{3}\)

Explanation:

\(\tan t = \dfrac{3}{7}\) : लम्ब \(= 3\), आधार \(= 7\), कर्ण \(= \sqrt{9 + 49} = \sqrt{58}\)
\(\operatorname{cosec}t = \dfrac{\sqrt{58}}{3}\) (कर्ण \(\div\) लम्ब)

\(\operatorname{cosec}(65^\circ + \theta) – \sec(25^\circ – \theta) + \tan^2 20^\circ – \operatorname{cosec}^2 70^\circ = ?\)

\(2\)
\(-1\)
\(1\)
\(0\)

Explanation:

\((65^\circ + \theta) + (25^\circ – \theta) = 90^\circ\), अतः \(\operatorname{cosec}(65^\circ+\theta) = \sec(25^\circ-\theta)\) और इनका अंतर \(0\) है।
\(\tan^2 20^\circ – \operatorname{cosec}^2 70^\circ = \tan^2 20^\circ – \sec^2 20^\circ = -1\)
मान \(= -1\)

यदि \(\dfrac{1 + \sin\theta}{1 – \sin\theta} = \dfrac{p^2}{q^2}\) है, तो \(\sec\theta\) निम्नलिखित में से किसके बराबर है:

\(\dfrac{2p^2q^2}{p^2 + q^2}\)
\(\dfrac{1}{2}\left(\dfrac{q}{p} + \dfrac{p}{q}\right)\)
\(\dfrac{1}{p^2} + \dfrac{1}{q^2}\)
\(\dfrac{p^2q^2}{p^2 + q^2}\)

Explanation:

\(\dfrac{1+\sin\theta}{1-\sin\theta} = \dfrac{(1+\sin\theta)^2}{\cos^2\theta} = (\sec\theta + \tan\theta)^2 = \dfrac{p^2}{q^2}\)
\(\sec\theta + \tan\theta = \dfrac{p}{q} \Rightarrow \sec\theta – \tan\theta = \dfrac{q}{p}\)
जोड़ने पर \(\sec\theta = \dfrac{1}{2}\left(\dfrac{p}{q} + \dfrac{q}{p}\right)\)

यदि \(\sec\theta + \tan\theta = p\), \((p > 1)\) तो \(\dfrac{\operatorname{cosec}\theta + 1}{\operatorname{cosec}\theta – 1} = ?\)

\(\dfrac{p + 1}{p – 1}\)
\(p^2\)
\(\dfrac{p – 1}{p + 1}\)
\(2p^2\)

Explanation:

\(\dfrac{\operatorname{cosec}\theta + 1}{\operatorname{cosec}\theta – 1} = \dfrac{1 + \sin\theta}{1 – \sin\theta}\) (अंश व हर को \(\sin\theta\) से गुणा करने पर)
\(\dfrac{1+\sin\theta}{1-\sin\theta} = (\sec\theta + \tan\theta)^2 = p^2\)

\(\Delta PQR\) में, \(\angle Q = 90^\circ\) है। यदि \(\cot R = \dfrac{1}{3}\) है, तो \(\dfrac{\sec P(\cos R + \sin P)}{\operatorname{cosec}R(\sin R – \operatorname{cosec}P)}\) का मान ज्ञात कीजिए।

\(\dfrac{2}{7}\)
\(-\dfrac{2}{7}\)
\(\dfrac{2}{3}\)
\(-\dfrac{2}{3}\)

Explanation:

\(\cot R = \dfrac{1}{3}\) : \(QR = k,\ PQ = 3k,\ PR = \sqrt{10}k\)
\(\sin P = \dfrac{QR}{PR} = \dfrac{1}{\sqrt{10}},\ \cos P = \dfrac{3}{\sqrt{10}}\)
\(\cos R = \dfrac{1}{\sqrt{10}},\ \sin R = \dfrac{3}{\sqrt{10}}\)
\(\sec P = \dfrac{\sqrt{10}}{3},\ \operatorname{cosec}P = \sqrt{10},\ \operatorname{cosec}R = \dfrac{\sqrt{10}}{3}\)
अंश \(= \dfrac{\sqrt{10}}{3}\left(\dfrac{1}{\sqrt{10}} + \dfrac{1}{\sqrt{10}}\right) = \dfrac{2}{3}\)
हर \(= \dfrac{\sqrt{10}}{3}\left(\dfrac{3}{\sqrt{10}} – \sqrt{10}\right) = \dfrac{3 – 10}{3} = -\dfrac{7}{3}\)
मान \(= \dfrac{2/3}{-7/3} = -\dfrac{2}{7}\)

यदि \(\dfrac{\sec\theta + \tan\theta}{\sec\theta – \tan\theta} = 2\dfrac{51}{79}\) है, तो \(\sin\theta\) का मान ज्ञात कीजिए।

\(\dfrac{91}{144}\)
\(\dfrac{39}{72}\)
\(\dfrac{65}{144}\)
\(\dfrac{35}{72}\)

Explanation:

\(\dfrac{\sec\theta + \tan\theta}{\sec\theta – \tan\theta} = \dfrac{1+\sin\theta}{1-\sin\theta} = 2\dfrac{51}{79} = \dfrac{209}{79}\)
\(79 + 79\sin\theta = 209 – 209\sin\theta \Rightarrow 288\sin\theta = 130\)
\(\sin\theta = \dfrac{65}{144}\)

यदि \(\cos\theta = \dfrac{5}{13}\) है, तो \(\tan^2\theta + \sec^2\theta\) का मान ज्ञात कीजिए।

\(\dfrac{233}{25}\)
\(\dfrac{303}{25}\)
\(\dfrac{313}{25}\)
\(\dfrac{323}{25}\)

Explanation:

\(\cos\theta = \dfrac{5}{13} \Rightarrow \sec^2\theta = \dfrac{169}{25}\)
\(\tan^2\theta = \sec^2\theta – 1 = \dfrac{144}{25}\)
\(\tan^2\theta + \sec^2\theta = \dfrac{144 + 169}{25} = \dfrac{313}{25}\)

यदि \(\operatorname{cosec}39^\circ = x\), तो \(\dfrac{1}{\operatorname{cosec}^2 51^\circ} + \sin^2 39^\circ + \tan^2 51^\circ – \dfrac{1}{\sin^2 51^\circ \sec^2 39^\circ}\) का मान ज्ञात कीजिए।

\(1 – x^2\)
\(\sqrt{x^2 – 1}\)
\(x^2 – 1\)
\(\sqrt{1 – x^2}\)

Explanation:

\(\sin 51^\circ = \cos 39^\circ,\ \tan 51^\circ = \cot 39^\circ\)
\(\dfrac{1}{\operatorname{cosec}^2 51^\circ} = \sin^2 51^\circ = \cos^2 39^\circ\)
\(\dfrac{1}{\sin^2 51^\circ\sec^2 39^\circ} = \dfrac{1}{\cos^2 39^\circ\sec^2 39^\circ} = 1\)
व्यंजक \(= \cos^2 39^\circ + \sin^2 39^\circ + \cot^2 39^\circ – 1 = \cot^2 39^\circ\)
\(\cot^2 39^\circ = \operatorname{cosec}^2 39^\circ – 1 = x^2 – 1\)

\(3(\cot^2 47^\circ – \sec^2 43^\circ) – \dfrac{2}{\tan^2 23^\circ – \operatorname{cosec}^2 67^\circ} + \operatorname{cosec}^2(68^\circ + \theta) – \tan(\theta + 61^\circ) – \tan^2(22^\circ – \theta) + \cot(29^\circ – \theta)\) का मान है:

\(1\)
\(0\)
\(5\)
\(-1\)

Explanation:

\(\cot 47^\circ\) और \(\sec 43^\circ = \operatorname{cosec}47^\circ\): \(3(\cot^2 47^\circ – \operatorname{cosec}^2 47^\circ) = -3\)
\(\operatorname{cosec}67^\circ = \sec 23^\circ\): \(\tan^2 23^\circ – \sec^2 23^\circ = -1\), अतः \(-\dfrac{2}{-1} = +2\)
\(\tan(22^\circ – \theta) = \cot(68^\circ+\theta)\): \(\operatorname{cosec}^2(68^\circ+\theta) – \cot^2(68^\circ+\theta) = 1\)
\(\tan(\theta+61^\circ) = \cot(29^\circ – \theta)\): \(-\tan(\theta+61^\circ) + \cot(29^\circ-\theta) = 0\)
कुल \(= -3 + 2 + 1 + 0 = 0\)
नोट: मैंने \(\theta\) के अलग-अलग मान रखकर संख्यात्मक जाँच भी की; मान हर बार \(0\) आता है (विकल्प 2)। कुंजी में \(-1\) दिया है, जो सही नहीं है। कृपया कुंजी जाँच लें।

यदि \(\sin A = \dfrac{5}{13}\) और \(7\cot B = 24\) है, तो \((\sec A \cos B)(\operatorname{cosec}B \tan A)\) का मान क्या होगा?

\(\dfrac{13}{7}\)
\(\dfrac{65}{42}\)
\(\dfrac{15}{13}\)
\(\dfrac{13}{14}\)

Explanation:

\(\sin A = \dfrac{5}{13} \Rightarrow \cos A = \dfrac{12}{13},\ \sec A = \dfrac{13}{12},\ \tan A = \dfrac{5}{12}\)
\(7\cot B = 24 \Rightarrow \tan B = \dfrac{7}{24} \Rightarrow \sin B = \dfrac{7}{25},\ \cos B = \dfrac{24}{25},\ \operatorname{cosec}B = \dfrac{25}{7}\)
\(\sec A\cos B = \dfrac{13}{12}\times\dfrac{24}{25} = \dfrac{26}{25}\)
\(\operatorname{cosec}B\tan A = \dfrac{25}{7}\times\dfrac{5}{12} = \dfrac{125}{84}\)
गुणनफल \(= \dfrac{26}{25}\times\dfrac{125}{84} = \dfrac{130}{84} = \dfrac{65}{42}\)

\(\dfrac{3(\operatorname{cosec}^2 26^\circ – \tan^2 64^\circ) + (\cot^2 42^\circ – \sec^2 48^\circ)}{\cot(22^\circ – \theta) – \operatorname{cosec}^2(62^\circ + \theta) – \tan(\theta + 68^\circ) + \tan^2(28^\circ – \theta)}\) का मान है:

\(3\)
\(4\)
\(-1\)
\(-2\)

Explanation:

\(\tan 64^\circ = \cot 26^\circ\): \(\operatorname{cosec}^2 26^\circ – \cot^2 26^\circ = 1\), अतः \(3(\ldots) = 3\)
\(\sec 48^\circ = \operatorname{cosec}42^\circ\): \(\cot^2 42^\circ – \operatorname{cosec}^2 42^\circ = -1\)
अंश \(= 3 – 1 = 2\)
हर में \(\cot(22^\circ – \theta) = \tan(68^\circ + \theta)\), अतः \(\cot(22^\circ-\theta) – \tan(\theta+68^\circ) = 0\)
\(\tan(28^\circ-\theta) = \cot(62^\circ+\theta)\): \(-\operatorname{cosec}^2(62^\circ+\theta) + \cot^2(62^\circ+\theta) = -1\)
हर \(= -1\)
मान \(= \dfrac{2}{-1} = -2\)

यदि \(\dfrac{\sec\theta + \tan\theta}{\sec\theta – \tan\theta} = \dfrac{1}{7}\), \(\theta\) प्रथम चतुर्थांश में स्थित है, तो \(\dfrac{\operatorname{cosec}\theta + \cot^2\theta}{\operatorname{cosec}\theta – \cot^2\theta} = ?\)

\(\dfrac{19}{5}\)
\(\dfrac{22}{3}\)
\(\dfrac{37}{12}\)
\(\dfrac{37}{19}\)

Explanation:

मानक रूप में \(\dfrac{\sec\theta + \tan\theta}{\sec\theta – \tan\theta} = \dfrac{1+\sin\theta}{1-\sin\theta}\)
यहाँ यह \(\dfrac{1}{7}\) देने पर \(\sin\theta = -\dfrac{3}{4}\) आता है, जो प्रथम चतुर्थांश में संभव नहीं है। कुंजी \(\dfrac{19}{5}\) तब आती है जब अनुपात \(7\) हो: \(\dfrac{1+\sin\theta}{1-\sin\theta} = 7 \Rightarrow \sin\theta = \dfrac{3}{4}\)
\(\operatorname{cosec}\theta = \dfrac{4}{3},\ \cot^2\theta = \operatorname{cosec}^2\theta – 1 = \dfrac{7}{9}\)
\(\dfrac{\frac{4}{3} + \frac{7}{9}}{\frac{4}{3} – \frac{7}{9}} = \dfrac{19/9}{5/9} = \dfrac{19}{5}\)
नोट: प्रश्न में अनुपात \(\frac{1}{7}\) की जगह \(7\) होना चाहिए (या अंश-हर उलटे), उसी के अनुसार हल किया गया है।

\(\dfrac{(\cos 9^\circ + \sin 81^\circ)(\sec 9^\circ + \operatorname{cosec}81^\circ)}{\operatorname{cosec}^2 71^\circ + \tan^2(90^\circ – 71^\circ) + 1}\) का मान है:

\(1\)
\(4\)
\(-3\)
\(2\)

Explanation:

अंश: \(\cos 9^\circ = \sin 81^\circ\) और \(\sec 9^\circ = \operatorname{cosec}81^\circ\)
\((2\cos 9^\circ)(2\sec 9^\circ) = 4\)
हर: \(\tan(90^\circ – 71^\circ) = \cot 71^\circ\); कुंजी के अनुसार हर \(\operatorname{cosec}^2 71^\circ – \cot^2 71^\circ + 1 = 2\)
मान \(= \dfrac{4}{2} = 2\)
नोट: प्रश्न में हर का चिह्न \(+\) छपा है; वैसा लेने पर मान स्थिर नहीं आता (लगभग \(1.79\))। कुंजी \(2\) तब आती है जब \(\tan^2\) से पहले \(-\) हो, इसलिए उसी के अनुसार हल किया गया है।

निम्नलिखित का मान निकालिए। \(\sin 25^\circ \sin 65^\circ – \cos 25^\circ \cos 65^\circ\)

\(4\)
\(1\)
\(0\)
\(40\)

Explanation:

\(\sin 65^\circ = \cos 25^\circ\) और \(\cos 65^\circ = \sin 25^\circ\)
\(\sin 25^\circ\cos 25^\circ – \cos 25^\circ\sin 25^\circ = 0\)

यदि \(\sin A = \dfrac{8}{17}\), तो \(\cot A + \sec A\) का मान क्या है?

\(2\dfrac{1}{120}\)
\(3\dfrac{1}{120}\)
\(4\dfrac{1}{120}\)
\(5\dfrac{1}{120}\)

Explanation:

\(\sin A = \dfrac{8}{17} \Rightarrow \cos A = \dfrac{15}{17}\)
\(\cot A = \dfrac{15}{8},\ \sec A = \dfrac{17}{15}\)
\(\cot A + \sec A = \dfrac{225 + 136}{120} = \dfrac{361}{120} = 3\dfrac{1}{120}\)

यदि \(4 – 2\sin^2\theta – 5\cos\theta = 0\), \(0^\circ < \theta < 90^\circ\), तो \(\sin\theta + \tan\theta\) का मान है:

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

Explanation:

\(\sin^2\theta = 1 – \cos^2\theta\) रखने पर
\(4 – 2(1 – \cos^2\theta) – 5\cos\theta = 0 \Rightarrow 2\cos^2\theta – 5\cos\theta + 2 = 0\)
\((2\cos\theta – 1)(\cos\theta – 2) = 0 \Rightarrow \cos\theta = \dfrac{1}{2}\) (क्योंकि \(\cos\theta \le 1\))
\(\theta = 60^\circ\)
\(\sin\theta + \tan\theta = \dfrac{\sqrt{3}}{2} + \sqrt{3} = \dfrac{3\sqrt{3}}{2}\)

\(\dfrac{2 + \tan^2\theta + \cot^2\theta}{\sec\theta \operatorname{cosec}\theta} = ?\)

\(\cot\theta\)
\(\cos\theta \cdot \sin\theta\)
\(\sec\theta \cdot \operatorname{cosec}\theta\)
\(\tan\theta\)

Explanation:

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

यदि \(\cos\theta = 2p/(1 + p^2)\), तो \(\tan\theta\) बराबर है:

\(\dfrac{p^2}{1 + p^2}\)
\(\dfrac{2p}{1 + p^2}\)
\(\dfrac{1 – p^2}{1 + p^2}\)
\(\dfrac{1 – p^2}{2p}\)

Explanation:

\(\cos\theta = \dfrac{2p}{1+p^2} \Rightarrow \sin\theta = \sqrt{1 – \dfrac{4p^2}{(1+p^2)^2}} = \dfrac{1 – p^2}{1 + p^2}\) (\(p \lt 1\))
\(\tan\theta = \dfrac{\sin\theta}{\cos\theta} = \dfrac{1 – p^2}{2p}\)

जब \(\sqrt{3}\cos\theta + \sin\theta = 1\) \((0^\circ \le \theta \le 90^\circ)\) है, तो \(\theta\) का मान है:

\(90^\circ\)
\(30^\circ\)
\(60^\circ\)
\(0^\circ\)

Explanation:

\(\sqrt{3}\cos\theta + \sin\theta = 2\left(\dfrac{\sqrt{3}}{2}\cos\theta + \dfrac{1}{2}\sin\theta\right) = 2\sin(\theta + 60^\circ) = 1\)
\(\sin(\theta + 60^\circ) = \dfrac{1}{2} \Rightarrow \theta + 60^\circ = 150^\circ\) (क्योंकि \(60^\circ \le \theta + 60^\circ \le 150^\circ\))
\(\theta = 90^\circ\)
जाँच: \(\sqrt{3}\times 0 + 1 = 1\)

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

\(\sec\theta \cdot \sin\theta\)
\(\sec\theta \cdot \tan\theta\)
\(\sec\theta + \tan\theta\)
\(\sec\theta – \tan\theta\)

Explanation:

अंश व हर को \(\cos\theta\) से भाग देने पर \(\dfrac{\tan\theta – 1 + \sec\theta}{\tan\theta + 1 – \sec\theta}\)
अंश में \(1 = \sec^2\theta – \tan^2\theta\) रखने पर अंश \(= (\sec\theta + \tan\theta) – (\sec\theta – \tan\theta)(\sec\theta+\tan\theta) = (\sec\theta + \tan\theta)(1 – \sec\theta + \tan\theta)\)
हर \(= \tan\theta + 1 – \sec\theta\)
मान \(= \sec\theta + \tan\theta\)

यदि \(12\cot^2\theta – 31\operatorname{cosec}\theta + 32 = 0\), \(0^\circ < \theta < 90^\circ\), तो \(\tan\theta\) के मान हैं:

\(\dfrac{4}{3}, \dfrac{3\sqrt{7}}{7}\)
\(\dfrac{4}{5}, \dfrac{5\sqrt{7}}{7}\)
\(\dfrac{5}{4}, \dfrac{4}{3}\)
\(\dfrac{4}{5}, \dfrac{4}{3}\)

Explanation:

\(\cot^2\theta = \operatorname{cosec}^2\theta – 1\); माना \(\operatorname{cosec}\theta = c\)
\(12(c^2 – 1) – 31c + 32 = 0 \Rightarrow 12c^2 – 31c + 20 = 0\)
\(c = \dfrac{31 \pm 1}{24} = \dfrac{4}{3}\) या \(\dfrac{5}{4}\)
\(c = \dfrac{4}{3}\): \(\sin\theta = \dfrac{3}{4},\ \tan\theta = \dfrac{3}{\sqrt{7}} = \dfrac{3\sqrt{7}}{7}\)
\(c = \dfrac{5}{4}\): \(\sin\theta = \dfrac{4}{5},\ \tan\theta = \dfrac{4}{3}\)
अतः \(\tan\theta = \dfrac{4}{3},\ \dfrac{3\sqrt{7}}{7}\)

यदि \(\dfrac{\cos^2\theta}{\cot^2\theta – \cos^2\theta} = 3\), \(0^\circ < \theta < 90^\circ\) है, तो \(\cot\theta + \operatorname{cosec}\theta\) का मान है।

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

Explanation:

\(\cot^2\theta – \cos^2\theta = \cos^2\theta\left(\dfrac{1}{\sin^2\theta} – 1\right) = \dfrac{\cos^4\theta}{\sin^2\theta}\)
\(\dfrac{\cos^2\theta}{\cos^4\theta/\sin^2\theta} = \tan^2\theta = 3 \Rightarrow \tan\theta = \sqrt{3},\ \theta = 60^\circ\)
\(\cot\theta + \operatorname{cosec}\theta = \dfrac{1}{\sqrt{3}} + \dfrac{2}{\sqrt{3}} = \sqrt{3}\)

\(\dfrac{1}{\sin\theta} – \dfrac{\cot^2\theta}{1 + \operatorname{cosec}\theta} = ?\)

\(0\)
\(1\)
\(2\)
\(-1\)

Explanation:

\(\cot^2\theta = \operatorname{cosec}^2\theta – 1 = (\operatorname{cosec}\theta – 1)(\operatorname{cosec}\theta + 1)\)
\(\dfrac{\cot^2\theta}{1 + \operatorname{cosec}\theta} = \operatorname{cosec}\theta – 1\)
मान \(= \operatorname{cosec}\theta – (\operatorname{cosec}\theta – 1) = 1\)

यदि \(\dfrac{\cos\theta}{1 – \sin\theta} + \dfrac{\cos\theta}{1 + \sin\theta} = 4\), \(0^\circ < \theta < 90^\circ\), तो \((\tan\theta + \operatorname{cosec}\theta)\) का मान है:

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

Explanation:

\(\dfrac{\cos\theta}{1-\sin\theta} + \dfrac{\cos\theta}{1+\sin\theta} = \dfrac{2\cos\theta}{1 – \sin^2\theta} = \dfrac{2}{\cos\theta} = 4\)
\(\cos\theta = \dfrac{1}{2} \Rightarrow \theta = 60^\circ\)
\(\tan\theta + \operatorname{cosec}\theta = \sqrt{3} + \dfrac{2}{\sqrt{3}} = \dfrac{5}{\sqrt{3}} = \dfrac{5\sqrt{3}}{3}\)

\(\left(\dfrac{1}{1 + \sin^2\theta} + \dfrac{1}{1 + \operatorname{cosec}^2\theta}\right) = ?\)

\(\sin^2\theta\)
\(1\)
\(\operatorname{cosec}^2\theta\)
\(2\)

Explanation:

\(1 + \operatorname{cosec}^2\theta = 1 + \dfrac{1}{\sin^2\theta} = \dfrac{\sin^2\theta + 1}{\sin^2\theta}\)
\(\dfrac{1}{1+\operatorname{cosec}^2\theta} = \dfrac{\sin^2\theta}{1 + \sin^2\theta}\)
योग \(= \dfrac{1 + \sin^2\theta}{1 + \sin^2\theta} = 1\)

\(\dfrac{1}{\sec x – \tan x} – \dfrac{1}{\cos x}\), \(0^\circ < x < 90^\circ\), का मान बराबर है:

\(2\sec x\)
\(\tan x\)
\(2\cos x\)
\(\cot x\)

Explanation:

\(\dfrac{1}{\sec x – \tan x} = \sec x + \tan x\) (क्योंकि \(\sec^2 x – \tan^2 x = 1\))
\(\sec x + \tan x – \dfrac{1}{\cos x} = \sec x + \tan x – \sec x = \tan x\)

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

\(\dfrac{5\sqrt{3}}{6}\)
\(\dfrac{5\sqrt{3}}{3}\)
\(2\sqrt{3}\)
\(3\sqrt{3}\)

Explanation:

\(\tan^2\theta = \sec^2\theta – 1\) रखने पर \(\sec^2\theta – 3\sec\theta + 2 = 0\)
\((\sec\theta – 1)(\sec\theta – 2) = 0 \Rightarrow \sec\theta = 2\) (\(\theta \ne 0^\circ\))
\(\theta = 60^\circ\)
\(\sin\theta + \cot\theta = \dfrac{\sqrt{3}}{2} + \dfrac{1}{\sqrt{3}} = \dfrac{3 + 2}{2\sqrt{3}} = \dfrac{5}{2\sqrt{3}} = \dfrac{5\sqrt{3}}{6}\)

यदि \(\cos^2\theta – 3\cos\theta + 2 = \sin^2\theta\), \(0^\circ < \theta < 90^\circ\), तो \(2\operatorname{cosec}\theta + 4\cot\theta\) का मान है:

\(\dfrac{8\sqrt{3}}{3}\)
\(\dfrac{4\sqrt{3}}{4}\)
\(2\sqrt{3}\)
\(4\sqrt{3}\)

Explanation:

\(\sin^2\theta = 1 – \cos^2\theta\) रखने पर \(\cos^2\theta – 3\cos\theta + 2 = 1 – \cos^2\theta\)
\(2\cos^2\theta – 3\cos\theta + 1 = 0 \Rightarrow (2\cos\theta – 1)(\cos\theta – 1) = 0\)
\(\cos\theta = \dfrac{1}{2}\) (क्योंकि \(\theta \ne 0^\circ\)), अतः \(\theta = 60^\circ\)
\(2\operatorname{cosec}\theta + 4\cot\theta = \dfrac{4}{\sqrt{3}} + \dfrac{4}{\sqrt{3}} = \dfrac{8}{\sqrt{3}} = \dfrac{8\sqrt{3}}{3}\)

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

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

Explanation:

\(\dfrac{1}{\operatorname{cosec}\theta – 1} + \dfrac{1}{\operatorname{cosec}\theta + 1} = \dfrac{2\operatorname{cosec}\theta}{\operatorname{cosec}^2\theta – 1} = \dfrac{2\operatorname{cosec}\theta}{\cot^2\theta}\)
\(\dfrac{2\operatorname{cosec}\theta}{\cot^2\theta} = 2\sec\theta \Rightarrow \dfrac{1}{\sin\theta}\cdot\dfrac{\sin^2\theta}{\cos^2\theta} = \dfrac{1}{\cos\theta} \Rightarrow \sin\theta = \cos\theta\)
\(\theta = 45^\circ\)
\(\cot 45^\circ + \cos 45^\circ = 1 + \dfrac{1}{\sqrt{2}} = \dfrac{2 + \sqrt{2}}{2}\)

यदि \(\dfrac{\tan\theta + \sin\theta}{\tan\theta – \sin\theta} = \dfrac{k + 1}{k – 1}\) है, तो \(k = ?\)

\(\operatorname{cosec}\theta\)
\(\sec\theta\)
\(\cos\theta\)
\(\sin\theta\)

Explanation:

अंश व हर को \(\sin\theta\) से भाग देने पर \(\dfrac{\tan\theta + \sin\theta}{\tan\theta – \sin\theta} = \dfrac{\sec\theta + 1}{\sec\theta – 1}\)
इसकी तुलना \(\dfrac{k+1}{k-1}\) से करने पर \(k = \sec\theta\)

\(\cot^2 62^\circ – \sec^2 28^\circ + \operatorname{cosec}^2 30^\circ + \tan^2 60^\circ = ?\)

\(8\)
\(\dfrac{10}{3}\)
\(6\)
\(\dfrac{16}{3}\)

Explanation:

\(\cot 62^\circ = \tan 28^\circ\): \(\cot^2 62^\circ – \sec^2 28^\circ = \tan^2 28^\circ – \sec^2 28^\circ = -1\)
\(\operatorname{cosec}^2 30^\circ = 4,\ \tan^2 60^\circ = 3\)
मान \(= -1 + 4 + 3 = 6\)

\(\sec^2 29^\circ – \cot^2 61^\circ + \sin^2 60^\circ + \operatorname{cosec}^2 30^\circ\) बराबर है:

\(\dfrac{19}{4}\)
\(\dfrac{23}{4}\)
\(\dfrac{15}{4}\)
\(\dfrac{11}{4}\)

Explanation:

\(\cot 61^\circ = \tan 29^\circ\): \(\sec^2 29^\circ – \tan^2 29^\circ = 1\)
\(\sin^2 60^\circ = \dfrac{3}{4},\ \operatorname{cosec}^2 30^\circ = 4\)
मान \(= 1 + \dfrac{3}{4} + 4 = \dfrac{23}{4}\)

\(\sin^2 32^\circ + \sin^2 58^\circ – \sin 30^\circ + \sec^2 60^\circ = ?\)

5.5
3.5
4.5
4

Explanation:

\(\sin 58^\circ = \cos 32^\circ\), अतः \(\sin^2 32^\circ + \sin^2 58^\circ = 1\)
\(\sin 30^\circ = \dfrac{1}{2},\ \sec^2 60^\circ = 4\)
मान \(= 1 – \dfrac{1}{2} + 4 = 4.5\)

यदि \(3\cos^2 A + 7\sin^2 A = 4\) है और A न्यून कोण (acute angle) दिया गया है, तो \(\cot A\) का मान होगा:

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

Explanation:

\(3\cos^2 A + 7\sin^2 A = 3(\cos^2 A + \sin^2 A) + 4\sin^2 A = 3 + 4\sin^2 A = 4\)
\(\sin^2 A = \dfrac{1}{4} \Rightarrow \sin A = \dfrac{1}{2} \Rightarrow A = 30^\circ\)
\(\cot 30^\circ = \sqrt{3}\)

\(\left[\dfrac{\sin^2 25^\circ + \sin^2 65^\circ}{\cos^2 24^\circ + \cos^2 66^\circ} + \sin^2 61^\circ + \cos 61^\circ \sin 29^\circ\right]\) का मान होगा:

\(2\)
\(1\)
\(3\)
\(0\)

Explanation:

\(\sin 65^\circ = \cos 25^\circ\): \(\sin^2 25^\circ + \sin^2 65^\circ = \sin^2 25^\circ + \cos^2 25^\circ = 1\)
\(\cos 66^\circ = \sin 24^\circ\): \(\cos^2 24^\circ + \cos^2 66^\circ = 1\)
पहला भाग \(= 1\)
\(\sin 29^\circ = \cos 61^\circ\): \(\sin^2 61^\circ + \cos 61^\circ\sin 29^\circ = \sin^2 61^\circ + \cos^2 61^\circ = 1\)
कुल मान \(= 1 + 1 = 2\)

यदि \(3\cos^2 A + 7\sin^2 A = 3\), \(0^\circ \le A \le 90^\circ\), तो A का मान है:

\(30^\circ\)
\(0^\circ\)
\(45^\circ\)
\(90^\circ\)

Explanation:

\(3\cos^2 A + 7\sin^2 A = 3 + 4\sin^2 A = 3\)
\(\sin^2 A = 0 \Rightarrow A = 0^\circ\)

यदि \(2\sin\theta – 8\cos^2\theta + 5 = 0\), \(0^\circ < \theta < 90^\circ\), है, तो \((\tan 2\theta + \operatorname{cosec}2\theta)\) मान क्या है?

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

Explanation:

\(\cos^2\theta = 1 – \sin^2\theta\) रखने पर \(2\sin\theta – 8(1 – \sin^2\theta) + 5 = 0\)
\(8\sin^2\theta + 2\sin\theta – 3 = 0 \Rightarrow (2\sin\theta – 1)(4\sin\theta + 3) = 0\)
\(\sin\theta = \dfrac{1}{2} \Rightarrow \theta = 30^\circ\), अतः \(2\theta = 60^\circ\)
\(\tan 60^\circ + \operatorname{cosec}60^\circ = \sqrt{3} + \dfrac{2}{\sqrt{3}} = \dfrac{5}{\sqrt{3}} = \dfrac{5\sqrt{3}}{3}\)

यदि \(7\sin^2\theta – \cos^2\theta + 2\sin\theta = 2\), \(0^\circ < \theta < 90^\circ\), तो \(\dfrac{\sec 2\theta + \cot 2\theta}{\operatorname{cosec}2\theta + \tan 2\theta}\) का मान है:

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

Explanation:

\(\cos^2\theta = 1 – \sin^2\theta\) रखने पर \(7\sin^2\theta – (1 – \sin^2\theta) + 2\sin\theta – 2 = 0\)
\(8\sin^2\theta + 2\sin\theta – 3 = 0 \Rightarrow \sin\theta = \dfrac{1}{2},\ \theta = 30^\circ,\ 2\theta = 60^\circ\)
\(\sec 60^\circ + \cot 60^\circ = 2 + \dfrac{1}{\sqrt{3}} = \dfrac{2\sqrt{3}+1}{\sqrt{3}}\)
\(\operatorname{cosec}60^\circ + \tan 60^\circ = \dfrac{2}{\sqrt{3}} + \sqrt{3} = \dfrac{5}{\sqrt{3}}\)
मान \(= \dfrac{2\sqrt{3}+1}{5}\)

व्यंजक \(3\sec^2\theta \cdot \tan^2\theta + \tan^6\theta – \sec^6\theta\) का मान बताइए।

\(-2\)
\(1\)
\(2\)
\(-1\)

Explanation:

\(\sec^6\theta = (1 + \tan^2\theta)^3 = 1 + 3\tan^2\theta + 3\tan^4\theta + \tan^6\theta\)
\(3\sec^2\theta\tan^2\theta = 3\tan^2\theta(1 + \tan^2\theta) = 3\tan^2\theta + 3\tan^4\theta\)
व्यंजक \(= 3\tan^2\theta + 3\tan^4\theta + \tan^6\theta – 1 – 3\tan^2\theta – 3\tan^4\theta – \tan^6\theta = -1\)

\(\dfrac{\tan^2\theta – \sin^2\theta}{2 + \tan^2\theta + \cot^2\theta}\) का मान है:

\(\operatorname{cosec}^6\theta\)
\(\cos^4\theta\)
\(\sin^6\theta\)
\(\sec^4\theta\)

Explanation:

अंश: \(\tan^2\theta – \sin^2\theta = \tan^2\theta(1 – \cos^2\theta) = \tan^2\theta\sin^2\theta = \dfrac{\sin^4\theta}{\cos^2\theta}\)
हर: \(2 + \tan^2\theta + \cot^2\theta = (\tan\theta + \cot\theta)^2 = \dfrac{1}{\sin^2\theta\cos^2\theta}\)
मान \(= \dfrac{\sin^4\theta}{\cos^2\theta}\times\sin^2\theta\cos^2\theta = \sin^6\theta\)

यदि \(\sec\theta + \tan\theta = p\), \(0^\circ < \theta < 90^\circ\) है, तो \(\dfrac{p^2 – 1}{p^2 + 1}\) ________ के बराबर है।

\(\operatorname{cosec}\theta\)
\(\sin\theta\)
\(\cos\theta\)
\(2\operatorname{cosec}\theta\)

Explanation:

\(\sec\theta + \tan\theta = p \Rightarrow \sec\theta – \tan\theta = \dfrac{1}{p}\)
\(\sec\theta = \dfrac{p^2+1}{2p},\ \tan\theta = \dfrac{p^2-1}{2p}\)
\(\sin\theta = \dfrac{\tan\theta}{\sec\theta} = \dfrac{p^2-1}{p^2+1}\)

यदि \(7\cos^2\theta + 3\sin^2\theta = 6\), \(0^\circ < \theta < 90^\circ\) है, तो \(\dfrac{\cot^2 2\theta + \sec^2 2\theta}{\tan^2 2\theta – \sin^2 2\theta}\) का मान ________ हैं।

\(\dfrac{52}{27}\)
\(\dfrac{26}{15}\)
\(\dfrac{28}{27}\)
\(\dfrac{49}{45}\)

Explanation:

\(7\cos^2\theta + 3\sin^2\theta = 3 + 4\cos^2\theta = 6 \Rightarrow \cos^2\theta = \dfrac{3}{4}\)
\(\theta = 30^\circ,\ 2\theta = 60^\circ\)
\(\cot^2 60^\circ = \dfrac{1}{3},\ \sec^2 60^\circ = 4,\ \tan^2 60^\circ = 3,\ \sin^2 60^\circ = \dfrac{3}{4}\)
अंश \(= \dfrac{13}{3}\), हर \(= \dfrac{9}{4}\)
मान \(= \dfrac{13}{3}\times\dfrac{4}{9} = \dfrac{52}{27}\)

\(\dfrac{(\cos 9^\circ + \sin 81^\circ)(\sec 9^\circ + \operatorname{cosec}81^\circ)}{2\sin^2 63^\circ + 1 + 2\sin^2 27^\circ}\)

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

Explanation:

अंश: \(\cos 9^\circ = \sin 81^\circ\) और \(\sec 9^\circ = \operatorname{cosec}81^\circ\), अतः \((2\cos 9^\circ)(2\sec 9^\circ) = 4\)
हर: \(\sin 27^\circ = \cos 63^\circ\), अतः \(2(\sin^2 63^\circ + \cos^2 63^\circ) + 1 = 3\)
मान \(= \dfrac{4}{3}\)

यदि \(12\cos^2\theta – 2\sin^2\theta + 3\cos\theta = 3\), \(0^\circ < \theta < 90^\circ\) है, तो \(\dfrac{\operatorname{cosec}\theta + \sec\theta}{\tan\theta + \cot\theta}\) का मान क्या है?

\(\dfrac{1 + 2\sqrt{3}}{2}\)
\(\dfrac{2 + \sqrt{3}}{4}\)
\(\dfrac{4 + \sqrt{3}}{4}\)
\(\dfrac{1 + \sqrt{3}}{2}\)

Explanation:

\(\sin^2\theta = 1 – \cos^2\theta\) रखने पर \(14\cos^2\theta + 3\cos\theta – 5 = 0\)
\(\cos\theta = \dfrac{-3 \pm 17}{28} = \dfrac{1}{2}\) (धनात्मक मान) \(\Rightarrow \theta = 60^\circ\)
\(\dfrac{\operatorname{cosec}\theta + \sec\theta}{\tan\theta + \cot\theta} = \dfrac{\frac{2}{\sqrt{3}} + 2}{\sqrt{3} + \frac{1}{\sqrt{3}}} = \dfrac{2(1+\sqrt{3})/\sqrt{3}}{4/\sqrt{3}} = \dfrac{1 + \sqrt{3}}{2}\)

\(\dfrac{\operatorname{cosec}(78^\circ + \theta) – \sec(12^\circ – \theta) – \tan(67^\circ + \theta) + \cot(23^\circ – \theta)}{\tan 13^\circ \tan 37^\circ \tan 45^\circ \tan 53^\circ \tan 77^\circ}\) का मान क्या है?

\(1\)
\(0\)
\(2\)
\(-1\)

Explanation:

\((78^\circ + \theta) + (12^\circ – \theta) = 90^\circ\), अतः \(\operatorname{cosec}(78^\circ+\theta) = \sec(12^\circ-\theta)\)
\((67^\circ + \theta) + (23^\circ – \theta) = 90^\circ\), अतः \(\tan(67^\circ+\theta) = \cot(23^\circ-\theta)\)
अंश \(= 0\)
हर \(= (\tan 13^\circ\tan 77^\circ)(\tan 37^\circ\tan 53^\circ)\tan 45^\circ = 1\)
मान \(= 0\)

यदि \(11\sin^2\theta – \cos^2\theta + 4\sin\theta – 4 = 0\), \(0^\circ < \theta < 90^\circ\) है, तो \(\dfrac{\cos 2\theta + \cot 2\theta}{\sec 2\theta – \tan 2\theta}\) का मान क्या हैं?

\(\dfrac{12 + 7\sqrt{3}}{6}\)
\(\dfrac{12 + 5\sqrt{3}}{3}\)
\(\dfrac{10 + 5\sqrt{3}}{3}\)
\(\dfrac{10 + 7\sqrt{3}}{6}\)

Explanation:

\(\cos^2\theta = 1 – \sin^2\theta\) रखने पर \(12\sin^2\theta + 4\sin\theta – 5 = 0\)
\(\sin\theta = \dfrac{-4 \pm 16}{24} = \dfrac{1}{2} \Rightarrow \theta = 30^\circ,\ 2\theta = 60^\circ\)
\(\dfrac{\cos 60^\circ + \cot 60^\circ}{\sec 60^\circ – \tan 60^\circ} = \dfrac{\frac{1}{2} + \frac{1}{\sqrt{3}}}{2 – \sqrt{3}} = \dfrac{\sqrt{3} + 2}{2\sqrt{3}}(2 + \sqrt{3})\)
\(= \dfrac{(2+\sqrt{3})^2}{2\sqrt{3}} = \dfrac{7 + 4\sqrt{3}}{2\sqrt{3}} = \dfrac{12 + 7\sqrt{3}}{6}\)

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

\(\dfrac{2(2\sqrt{3} – 1)}{3}\)
\(\dfrac{4\sqrt{3} – 1}{6}\)
\(\dfrac{3\sqrt{3} – 1}{6}\)
\(\dfrac{2\sqrt{3} – 1}{3}\)

Explanation:

\(\tan\theta – \cot\theta = \dfrac{\sin^2\theta – \cos^2\theta}{\sin\theta\cos\theta} = \dfrac{1}{\sin\theta}\)
\(\sin^2\theta – \cos^2\theta = \cos\theta \Rightarrow 1 – 2\cos^2\theta = \cos\theta \Rightarrow 2\cos^2\theta + \cos\theta – 1 = 0\)
\(\cos\theta = \dfrac{1}{2} \Rightarrow \theta = 60^\circ\)
\(\dfrac{2\tan 60^\circ – \cos 60^\circ}{\sqrt{3}\cot 60^\circ + \sec 60^\circ} = \dfrac{2\sqrt{3} – \frac{1}{2}}{1 + 2} = \dfrac{4\sqrt{3} – 1}{6}\)

यदि \((\cos^2\theta – 1)(1 + \tan^2\theta) + 2\tan^2\theta = 1\), \(0^\circ \le \theta \le 90^\circ\) है तो \(\theta\) का मान होगा?

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

Explanation:

\((\cos^2\theta – 1)(1 + \tan^2\theta) = -\sin^2\theta\cdot\sec^2\theta = -\tan^2\theta\)
\(-\tan^2\theta + 2\tan^2\theta = 1 \Rightarrow \tan^2\theta = 1\)
\(\theta = 45^\circ\)

\(4\left[\dfrac{(1 – \sec A)^2 + (1 + \sec A)^2}{1 + \sec^2 A}\right] = ?\)

\(8\)
\(1\)
\(4\)
\(2\)

Explanation:

\((1 – \sec A)^2 + (1 + \sec A)^2 = 2 + 2\sec^2 A = 2(1 + \sec^2 A)\)
\(4\times\dfrac{2(1 + \sec^2 A)}{1 + \sec^2 A} = 8\)

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

\(-1\)
\(1\)
\(0\)
\(-\sin^2\theta\cos^2\theta\)

Explanation:

\(\sin^4\theta + \cos^4\theta = (\sin^2\theta + \cos^2\theta)^2 – 2\sin^2\theta\cos^2\theta = 1 – 2\sin^2\theta\cos^2\theta\)
अतः \(\dfrac{1 – 2\sin^2\theta\cos^2\theta}{\sin^4\theta + \cos^4\theta} = 1\)
मान \(= 1 – 1 = 0\)

यदि \(3\sec^2\theta + \tan\theta = 7\), \(0^\circ < \theta < 90^\circ\) है, तो \(\dfrac{\operatorname{cosec}2\theta + \cos\theta}{\sin 2\theta + \cot\theta}\) का मान ज्ञात कीजिए।

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

Explanation:

\(3\sec^2\theta + \tan\theta = 7 \Rightarrow 3 + 3\tan^2\theta + \tan\theta – 7 = 0 \Rightarrow 3\tan^2\theta + \tan\theta – 4 = 0\)
\((3\tan\theta + 4)(\tan\theta – 1) = 0 \Rightarrow \tan\theta = 1,\ \theta = 45^\circ,\ 2\theta = 90^\circ\)
\(\dfrac{\operatorname{cosec}90^\circ + \cos 45^\circ}{\sin 90^\circ + \cot 45^\circ} = \dfrac{1 + \frac{1}{\sqrt{2}}}{2} = \dfrac{2 + \sqrt{2}}{4}\)

यदि \(\cot\theta + \tan\theta = 2\sec\theta\), \(0^\circ < \theta < 90^\circ\), तो \(\dfrac{\tan 2\theta – \sec\theta}{\cot 2\theta + \operatorname{cosec}\theta}\) का मान क्या होगा?

\(\dfrac{3 – \sqrt{2}}{11}\)
\(\dfrac{2\sqrt{3} – 1}{5}\)
\(\dfrac{2\sqrt{3} – 1}{11}\)
\(\dfrac{3 – \sqrt{2}}{5}\)

Explanation:

\(\cot\theta + \tan\theta = \dfrac{1}{\sin\theta\cos\theta} = \dfrac{2}{\cos\theta} \Rightarrow \sin\theta = \dfrac{1}{2}\)
\(\theta = 30^\circ,\ 2\theta = 60^\circ\)
\(\dfrac{\tan 60^\circ – \sec 30^\circ}{\cot 60^\circ + \operatorname{cosec}30^\circ} = \dfrac{\sqrt{3} – \frac{2}{\sqrt{3}}}{\frac{1}{\sqrt{3}} + 2} = \dfrac{1/\sqrt{3}}{(1 + 2\sqrt{3})/\sqrt{3}} = \dfrac{1}{1 + 2\sqrt{3}}\)
\(= \dfrac{2\sqrt{3} – 1}{12 – 1} = \dfrac{2\sqrt{3} – 1}{11}\)

यदि \(5\cos^2\theta + 1 = 3\sin^2\theta\), \(0^\circ < \theta < 90^\circ\), तो \(\dfrac{\tan\theta + \sec\theta}{\cot\theta + \operatorname{cosec}\theta}\) का मान क्या होगा?

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

Explanation:

\(5\cos^2\theta + 1 = 3(1 – \cos^2\theta) \Rightarrow 8\cos^2\theta = 2 \Rightarrow \cos\theta = \dfrac{1}{2}\)
\(\theta = 60^\circ\)
\(\dfrac{\tan 60^\circ + \sec 60^\circ}{\cot 60^\circ + \operatorname{cosec}60^\circ} = \dfrac{\sqrt{3} + 2}{\frac{3}{\sqrt{3}}} = \dfrac{\sqrt{3} + 2}{\sqrt{3}} = \dfrac{3 + 2\sqrt{3}}{3}\)

यदि \(6\tan\theta – 5\sqrt{3}\sec\theta + 12\cot\theta = 0\), \(0^\circ < \theta < 90^\circ\) हैं, तो \((\operatorname{cosec}\theta + \sec\theta)\) का मान ज्ञात कीजिए।

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

Explanation:

दोनों पक्षों को \(\sin\theta\cos\theta\) से गुणा करने पर \(6\sin^2\theta – 5\sqrt{3}\sin\theta + 12\cos^2\theta = 0\)
\(\cos^2\theta = 1 – \sin^2\theta\) रखने पर \(6\sin^2\theta + 5\sqrt{3}\sin\theta – 12 = 0\)
\(\sin\theta = \dfrac{-5\sqrt{3} \pm 11\sqrt{3}}{12} = \dfrac{\sqrt{3}}{2} \Rightarrow \theta = 60^\circ\)
\(\operatorname{cosec}60^\circ + \sec 60^\circ = \dfrac{2}{\sqrt{3}} + 2 = \dfrac{2(\sqrt{3} + 3)}{3} = \dfrac{2}{3}(3 + \sqrt{3})\)

निम्नलिखित को हल करें: \(\dfrac{2\sin 22^\circ}{\cos 68^\circ} – \dfrac{2\cot 75^\circ}{5\tan 15^\circ} – \dfrac{8\tan 45^\circ \tan 20^\circ \tan 40^\circ \tan 50^\circ \tan 70^\circ}{5}\)

\(3\)
\(2\)
\(0\)
\(1\)

Explanation:

\(\cos 68^\circ = \sin 22^\circ\), अतः \(\dfrac{2\sin 22^\circ}{\cos 68^\circ} = 2\)
\(\cot 75^\circ = \tan 15^\circ\), अतः \(\dfrac{2\cot 75^\circ}{5\tan 15^\circ} = \dfrac{2}{5}\)
\(\tan 20^\circ\tan 70^\circ = 1,\ \tan 40^\circ\tan 50^\circ = 1,\ \tan 45^\circ = 1\), अतः तीसरा पद \(= \dfrac{8}{5}\)
मान \(= 2 – \dfrac{2}{5} – \dfrac{8}{5} = 0\)

\((\operatorname{cosec}A + \cot A + 1)(\operatorname{cosec}A – \cot A + 1) – 2\operatorname{cosec}A = ?\)

\(2\operatorname{cosec}A\)
\(0\)
\(2\)
\(4\operatorname{cosec}A\)

Explanation:

\((\operatorname{cosec}A + \cot A + 1)(\operatorname{cosec}A – \cot A + 1) = (\operatorname{cosec}A + 1)^2 – \cot^2 A\)
\(= \operatorname{cosec}^2 A + 2\operatorname{cosec}A + 1 – (\operatorname{cosec}^2 A – 1) = 2\operatorname{cosec}A + 2\)
मान \(= 2\operatorname{cosec}A + 2 – 2\operatorname{cosec}A = 2\)

\(\dfrac{3(1 – 2\sin^2 x)}{\cos^2 x – \sin^2 x} = ?\)

\(2\)
\(1\)
\(3\)
\(4\)

Explanation:

\(1 – 2\sin^2 x = \cos^2 x – \sin^2 x\) (क्योंकि \(1 = \sin^2 x + \cos^2 x\))
मान \(= \dfrac{3(\cos^2 x – \sin^2 x)}{\cos^2 x – \sin^2 x} = 3\)

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

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

Explanation:

\(\cot^2\theta + \sin^2\theta – 1 = \cot^2\theta – \cos^2\theta = \dfrac{\cos^4\theta}{\sin^2\theta}\)
\(\dfrac{\cos^2\theta\sin^2\theta}{\cos^4\theta} = \tan^2\theta = 3 \Rightarrow \theta = 60^\circ\)
\(\tan 60^\circ + \operatorname{cosec}60^\circ = \sqrt{3} + \dfrac{2}{\sqrt{3}} = \dfrac{5\sqrt{3}}{3}\)

\(\dfrac{\operatorname{cosec}\theta}{\operatorname{cosec}\theta – 1} + \dfrac{\operatorname{cosec}\theta}{\operatorname{cosec}\theta + 1} – \tan^2\theta\), \(0^\circ < \theta < 90^\circ\), का मान ज्ञात करें।

\(2\sec^2\theta\)
\(\sec^2\theta + 1\)
\(\sec^2\theta\)
\(1 – \tan^2\theta\)

Explanation:

\(\dfrac{\operatorname{cosec}\theta}{\operatorname{cosec}\theta – 1} + \dfrac{\operatorname{cosec}\theta}{\operatorname{cosec}\theta + 1} = \dfrac{2\operatorname{cosec}^2\theta}{\operatorname{cosec}^2\theta – 1} = \dfrac{2\operatorname{cosec}^2\theta}{\cot^2\theta} = 2\sec^2\theta\)
मान \(= 2\sec^2\theta – \tan^2\theta = \sec^2\theta + (\sec^2\theta – \tan^2\theta) = \sec^2\theta + 1\)

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

\(7/4\)
\(5/4\)
\(9/4\)
\(3/4\)

Explanation:

\(2(1 – \sin^2\theta) = 3\sin\theta \Rightarrow 2\sin^2\theta + 3\sin\theta – 2 = 0\)
\((2\sin\theta – 1)(\sin\theta + 2) = 0 \Rightarrow \sin\theta = \dfrac{1}{2},\ \theta = 30^\circ\)
\(\sec^2\theta – \tan^2\theta + \cos^2\theta = 1 + \dfrac{3}{4} = \dfrac{7}{4}\)

यदि \(3\tan\theta = 2\sqrt{3}\sin\theta\), \(0^\circ < \theta < 90^\circ\) है, तो \(2\sin^2 2\theta – 3\cos^2 3\theta\) का मान ज्ञात करें।

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

Explanation:

\(3\tan\theta = 2\sqrt{3}\sin\theta \Rightarrow \dfrac{3}{\cos\theta} = 2\sqrt{3} \Rightarrow \cos\theta = \dfrac{\sqrt{3}}{2}\)
\(\theta = 30^\circ\)
\(2\sin^2 60^\circ – 3\cos^2 90^\circ = 2\times\dfrac{3}{4} – 0 = \dfrac{3}{2}\)

यदि \(3\sec\theta + 4\cos\theta – 4\sqrt{3} = 0\) है, जहां \(\theta\) न्यून कोण है, तो \(\theta\) का मान ज्ञात करें।

\(20^\circ\)
\(30^\circ\)
\(60^\circ\)
\(45^\circ\)

Explanation:

\(\dfrac{3}{\cos\theta} + 4\cos\theta = 4\sqrt{3} \Rightarrow 4\cos^2\theta – 4\sqrt{3}\cos\theta + 3 = 0\)
\((2\cos\theta – \sqrt{3})^2 = 0 \Rightarrow \cos\theta = \dfrac{\sqrt{3}}{2}\)
\(\theta = 30^\circ\)

\(1 + 2\tan^2\theta + 2\sin\theta\sec^2\theta\), \(0^\circ < \theta < 90^\circ\) का मान ज्ञात करें।

\(\dfrac{1 – \sin\theta}{1 + \sin\theta}\)
\(\dfrac{1 + \cos\theta}{1 – \cos\theta}\)
\(\dfrac{1 – \cos\theta}{1 + \cos\theta}\)
\(\dfrac{1 + \sin\theta}{1 – \sin\theta}\)

Explanation:

\(1 + \tan^2\theta = \sec^2\theta\), अतः व्यंजक \(= \sec^2\theta + \tan^2\theta + 2\sin\theta\sec^2\theta\)
\(= \dfrac{1 + \sin^2\theta + 2\sin\theta}{\cos^2\theta} = \dfrac{(1 + \sin\theta)^2}{(1 – \sin\theta)(1 + \sin\theta)} = \dfrac{1 + \sin\theta}{1 – \sin\theta}\)

यदि \(3\cos^2\theta – 4\sin\theta + 1 = 0\), \(0^\circ < \theta < 90^\circ\) है, तो \(\tan\theta + \sec\theta\) कितना होगा?

\(2\sqrt{3}\)
\(2\sqrt{5}\)
\(3\sqrt{3}\)
\(\sqrt{5}\)

Explanation:

\(3(1 – \sin^2\theta) – 4\sin\theta + 1 = 0 \Rightarrow 3\sin^2\theta + 4\sin\theta – 4 = 0\)
\((3\sin\theta – 2)(\sin\theta + 2) = 0 \Rightarrow \sin\theta = \dfrac{2}{3}\)
\(\cos\theta = \dfrac{\sqrt{5}}{3}\)
\(\tan\theta + \sec\theta = \dfrac{\sin\theta + 1}{\cos\theta} = \dfrac{5/3}{\sqrt{5}/3} = \sqrt{5}\)

यदि \(\cos\theta – \sin\theta = \sqrt{3}\cos(90^\circ – \theta)\), \(0^\circ < \theta < 90^\circ\) है, तो \(\tan\theta – \cot\theta\) का मान ज्ञात करें।

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

Explanation:

\(\cos(90^\circ – \theta) = \sin\theta\), अतः \(\cos\theta – \sin\theta = \sqrt{3}\sin\theta \Rightarrow \cos\theta = (1 + \sqrt{3})\sin\theta\)
\(\tan\theta = \dfrac{1}{1 + \sqrt{3}} = \dfrac{\sqrt{3} – 1}{2},\ \cot\theta = 1 + \sqrt{3}\)
\(\tan\theta – \cot\theta = \dfrac{1}{1 + \sqrt{3}} – (1 + \sqrt{3}) = \dfrac{1 – (1+\sqrt{3})^2}{1 + \sqrt{3}} = \dfrac{1 – 4 – 2\sqrt{3}}{1 + \sqrt{3}}\)
\(= -\dfrac{3 + 2\sqrt{3}}{1 + \sqrt{3}}\)

\(0^\circ < \theta < 90^\circ\) के लिए, \(\dfrac{1}{\cos\theta} + \dfrac{1}{\tan\theta – \sec\theta}\) का मान ज्ञात करें।

\(-\sec\theta\)
\(\tan\theta\)
\(\sec\theta\)
\(-\tan\theta\)

Explanation:

\(\dfrac{1}{\tan\theta – \sec\theta} = -\dfrac{1}{\sec\theta – \tan\theta} = -(\sec\theta + \tan\theta)\)
मान \(= \sec\theta – \sec\theta – \tan\theta = -\tan\theta\)

यदि \(4\sin^2(2x – 10)^\circ = 3\), \(0^\circ \le (2x – 10)^\circ \le 90^\circ\) है, तो \(\dfrac{\sin^4(x – 5)^\circ + \cos^4(x – 5)^\circ}{1 – 2\sin^2(3x – 15)^\circ\cos^2(3x – 15)^\circ}\) का मान ज्ञात करें।

\(1\)
\(5/8\)
\(-5/8\)
\(-1\)

Explanation:

\(4\sin^2(2x-10)^\circ = 3 \Rightarrow \sin(2x-10)^\circ = \dfrac{\sqrt{3}}{2} \Rightarrow 2x – 10 = 60 \Rightarrow x = 35\)
\(x – 5 = 30\) और \(3x – 15 = 90\)
अंश \(= \sin^4 30^\circ + \cos^4 30^\circ = \dfrac{1}{16} + \dfrac{9}{16} = \dfrac{5}{8}\)
हर \(= 1 – 2\sin^2 90^\circ\cos^2 90^\circ = 1\)
मान \(= \dfrac{5}{8}\)

यदि \(7\cos^2\theta + 5\sin^2\theta – 6 = 0\), \((0^\circ < \theta < 90^\circ)\) है, तो \(\sqrt{\dfrac{\operatorname{cosec}\theta + \tan\theta}{\sec\theta – \cot\theta}}\) का मान ज्ञात करें।

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

Explanation:

\(7\cos^2\theta + 5\sin^2\theta = 5 + 2\cos^2\theta = 6 \Rightarrow \cos^2\theta = \dfrac{1}{2}\)
\(\theta = 45^\circ\)
\(\dfrac{\operatorname{cosec}45^\circ + \tan 45^\circ}{\sec 45^\circ – \cot 45^\circ} = \dfrac{\sqrt{2} + 1}{\sqrt{2} – 1} = (\sqrt{2} + 1)^2\)
वर्गमूल \(= \sqrt{2} + 1\)

\((\sec\theta + \tan\theta)^2 + \dfrac{1 + \operatorname{cosec}\theta}{1 – \operatorname{cosec}\theta}\) का मान ज्ञात करें। जहां \(0^\circ < \theta < 90^\circ\) है।

\(0\)
\(-2\)
\(1\)
\(2\)

Explanation:

\((\sec\theta + \tan\theta)^2 = \dfrac{(1 + \sin\theta)^2}{\cos^2\theta} = \dfrac{1 + \sin\theta}{1 – \sin\theta}\)
\(\dfrac{1 + \operatorname{cosec}\theta}{1 – \operatorname{cosec}\theta} = \dfrac{\sin\theta + 1}{\sin\theta – 1} = -\dfrac{1 + \sin\theta}{1 – \sin\theta}\)
योग \(= 0\)

यदि \(2\cos^2\theta – 5\cos\theta + 2 = 0\) है, \(0^\circ < \theta < 90^\circ\), तो \((\sec\theta + \tan\theta)\) का मान ज्ञात करें।

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

Explanation:

\(2\cos^2\theta – 5\cos\theta + 2 = 0 \Rightarrow (2\cos\theta – 1)(\cos\theta – 2) = 0\)
\(\cos\theta = \dfrac{1}{2} \Rightarrow \theta = 60^\circ\)
\(\sec 60^\circ + \tan 60^\circ = 2 + \sqrt{3}\)

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

\(\sqrt[4]{3}\)
\(\sqrt{2}\)
\(\sqrt{3}\)
\(\sqrt[4]{2}\)

Explanation:

\(\sin^2\theta – (1 – \sin^2\theta) – 3\sin\theta + 2 = 0 \Rightarrow 2\sin^2\theta – 3\sin\theta + 1 = 0\)
\((2\sin\theta – 1)(\sin\theta – 1) = 0\); \(\theta \ne 90^\circ\) अतः \(\sin\theta = \dfrac{1}{2},\ \theta = 30^\circ\)
\(\sec 30^\circ – \tan 30^\circ = \dfrac{2}{\sqrt{3}} – \dfrac{1}{\sqrt{3}} = \dfrac{1}{\sqrt{3}}\)
\(\dfrac{1}{\sqrt{1/\sqrt{3}}} = \sqrt{\sqrt{3}} = \sqrt[4]{3}\)

यदि \(\dfrac{1}{1 – \sin\theta} + \dfrac{1}{1 + \sin\theta} = 4\sec\theta\), \(0^\circ < \theta < 90^\circ\) है, तो \(\cot\theta + \operatorname{cosec}\theta\) का मान ज्ञात करें।

\(5\sqrt{3}/3\)
\(3\sqrt{3}\)
\(\sqrt{3}\)
\(4\sqrt{3}/3\)

Explanation:

\(\dfrac{1}{1 – \sin\theta} + \dfrac{1}{1 + \sin\theta} = \dfrac{2}{\cos^2\theta} = \dfrac{4}{\cos\theta} \Rightarrow \cos\theta = \dfrac{1}{2}\)
\(\theta = 60^\circ\)
\(\cot 60^\circ + \operatorname{cosec}60^\circ = \dfrac{1}{\sqrt{3}} + \dfrac{2}{\sqrt{3}} = \sqrt{3}\)

यदि \(\tan\theta + 3\cot\theta – 2\sqrt{3} = 0\), \(0^\circ < \theta < 90^\circ\) है, तो \((\operatorname{cosec}^2\theta + \cos^2\theta)\) का मान ज्ञात करें।

\(\dfrac{19}{12}\)
\(\dfrac{2}{3}\)
\(\dfrac{11}{12}\)
\(\dfrac{14}{3}\)

Explanation:

\(\tan\theta + \dfrac{3}{\tan\theta} = 2\sqrt{3} \Rightarrow \tan^2\theta – 2\sqrt{3}\tan\theta + 3 = 0\)
\((\tan\theta – \sqrt{3})^2 = 0 \Rightarrow \tan\theta = \sqrt{3},\ \theta = 60^\circ\)
\(\operatorname{cosec}^2 60^\circ + \cos^2 60^\circ = \dfrac{4}{3} + \dfrac{1}{4} = \dfrac{19}{12}\)

यदि \(\sin^2\theta = 2\sin\theta – 1\), \(0^\circ \le \theta \le 90^\circ\) है, तो \(\dfrac{1 + \operatorname{cosec}\theta}{1 – \cos\theta}\) का मान ज्ञात करें।

\(-1\)
\(1\)
\(2\)
\(-2\)

Explanation:

\(\sin^2\theta – 2\sin\theta + 1 = 0 \Rightarrow (\sin\theta – 1)^2 = 0 \Rightarrow \sin\theta = 1\)
\(\theta = 90^\circ\)
\(\dfrac{1 + \operatorname{cosec}90^\circ}{1 – \cos 90^\circ} = \dfrac{1 + 1}{1 – 0} = 2\)

हल करें: यदि \(0^\circ < \theta < 90^\circ\) \(\dfrac{(\sin\theta + \sec\theta)^2 + (\cos\theta + \operatorname{cosec}\theta)^2}{(1 + \sec\theta\operatorname{cosec}\theta)^2}\)

\(0\)
\(2\)
\(-1\)
\(1\)

Explanation:

\((\sin\theta + \sec\theta)^2 + (\cos\theta + \operatorname{cosec}\theta)^2 = (\sin^2\theta + \cos^2\theta) + 2(\tan\theta + \cot\theta) + \sec^2\theta + \operatorname{cosec}^2\theta\)
माना \(m = \sec\theta\operatorname{cosec}\theta = \dfrac{1}{\sin\theta\cos\theta}\); तब \(\tan\theta + \cot\theta = m\) और \(\sec^2\theta + \operatorname{cosec}^2\theta = m^2\)
अंश \(= 1 + 2m + m^2 = (1 + m)^2\)
मान \(= \dfrac{(1+m)^2}{(1+m)^2} = 1\)

यदि \(4\sin^2\theta = 3(1 + \cos\theta)\) है, \(0^\circ < \theta < 90^\circ\), तो \((2\tan\theta + 4\sin\theta – \sec\theta)\) का मान ज्ञात करें।

\(3\sqrt{15} – 4\)
\(15\sqrt{3} – 4\)
\(15\sqrt{3} + 4\)
\(4\sqrt{15} – 3\)

Explanation:

\(4(1 – \cos^2\theta) = 3(1 + \cos\theta) \Rightarrow 4(1 – \cos\theta) = 3\) (\(1 + \cos\theta\) से भाग देने पर)
\(\cos\theta = \dfrac{1}{4},\ \sin\theta = \dfrac{\sqrt{15}}{4},\ \tan\theta = \sqrt{15},\ \sec\theta = 4\)
\(2\tan\theta + 4\sin\theta – \sec\theta = 2\sqrt{15} + \sqrt{15} – 4 = 3\sqrt{15} – 4\)

यदि \(3\sec^2\theta + \tan\theta – 7 = 0\), \(0^\circ < \theta < 90^\circ\) तो \(\left(\dfrac{2\sin\theta + 3\cos\theta}{\operatorname{cosec}\theta + \sec\theta}\right)\) का मान ज्ञात कीजिए।

\(10\)
\(5/2\)
\(5/4\)
\(4\sqrt{2}\)

Explanation:

\(3\sec^2\theta + \tan\theta – 7 = 0 \Rightarrow 3\tan^2\theta + \tan\theta – 4 = 0 \Rightarrow \tan\theta = 1,\ \theta = 45^\circ\)
\(\dfrac{2\sin 45^\circ + 3\cos 45^\circ}{\operatorname{cosec}45^\circ + \sec 45^\circ} = \dfrac{5/\sqrt{2}}{2\sqrt{2}} = \dfrac{5}{4}\)

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

\(\cos^2\theta – 1\)
\(\sec^2\theta + 1\)
\(\cot^2\theta – 1\)
\(\sin^2\theta – 1\)

Explanation:

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

\(\tan^2 A + 5\sec A = 13\) है, जहां \(0 < A < 90^\circ\) है। A का माप (डिग्री में) बताइए।

\(0\)
\(60\)
\(45\)
\(30\)

Explanation:

\(\tan^2 A = \sec^2 A – 1\) रखने पर \(\sec^2 A + 5\sec A – 14 = 0\)
\((\sec A + 7)(\sec A – 2) = 0 \Rightarrow \sec A = 2\)
\(A = 60^\circ\)

यदि \(\sin^2\theta – \cos^2\theta – 3\sin\theta + 2 = 0\), \(0^\circ < \theta < 90^\circ\) तो \(1 + \sec\theta + \tan\theta\) का मान कितना होगा?

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

Explanation:

\(2\sin^2\theta – 3\sin\theta + 1 = 0 \Rightarrow (2\sin\theta – 1)(\sin\theta – 1) = 0\)
\(\theta \ne 90^\circ\), अतः \(\sin\theta = \dfrac{1}{2},\ \theta = 30^\circ\)
\(1 + \sec 30^\circ + \tan 30^\circ = 1 + \dfrac{2}{\sqrt{3}} + \dfrac{1}{\sqrt{3}} = 1 + \sqrt{3}\)

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

\(\cos^2\theta\)
\(1\)
\(\cot^2\theta\)
\(-1\)

Explanation:

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

यदि \(\sec^2\theta + \tan^2\theta = 3\dfrac{1}{2}\), \(0^\circ < \theta < 90^\circ\) तो, \((\cos\theta + \sin\theta)\) के बराबर है?

\(\dfrac{1 + \sqrt{5}}{3}\)
\(\dfrac{2 + \sqrt{5}}{3}\)
\(\dfrac{1 + \sqrt{5}}{6}\)
\(\dfrac{9 + 2\sqrt{5}}{6}\)

Explanation:

\(\sec^2\theta + \tan^2\theta = 1 + 2\tan^2\theta = \dfrac{7}{2} \Rightarrow \tan^2\theta = \dfrac{5}{4}\)
\(\cos^2\theta = \dfrac{1}{1 + \tan^2\theta} = \dfrac{4}{9} \Rightarrow \cos\theta = \dfrac{2}{3}\)
\(\sin\theta = \dfrac{\sqrt{5}}{3}\)
\(\cos\theta + \sin\theta = \dfrac{2 + \sqrt{5}}{3}\)