SSC CGL 202031)If \(11 \sin^{2} \theta - \cos^{2} \theta + 4 \sin \theta - 4 = 0, 0^\circ < \theta < 90^\circ\), then what is the value of \(\frac{\cos 2\theta + \cot 2 \theta}{\sec 2 \theta - \tan 2 \theta}\) ?
\(12+7\sqrt3\over6\)
\(11 \sin^{2} \theta - \cos^{2} \theta + 4 \sin \theta - 4 = 0\);
\(11 \sin^{2} \theta - (1 - \sin^{2} \theta) + 4 \sin \theta - 4 = 0\);
\(12 \sin^{2} \theta + 4 \sin \theta - 5 = 0\);
\(12 \sin^{2} \theta + 10 \sin \theta - 6\sin \theta- 5 = 0\);
\(2\sin \theta(6 \sin\theta + 5) -1(6 \sin\theta + 5) = 0\);
\((2\sin \theta - 1)(6 \sin\theta + 5) = 0\);
\(For 0^\circ < \theta < 90^\circ,\)
\(\sin \theta = 1/2\); \( \theta = 30^0\);
\(\frac{\cos 2\theta + \cot 2 \theta}{\sec 2 \theta - \tan 2 \theta}\)
On putting the value of \(\theta\),
\(\frac{\cos 2\times 30 + \cot 2 \times 30}{\sec 2 \times 30 - \tan 2 \times 30}\); ⇒ \(\frac{\cos 60 + \cot 60}{\sec60 - \tan 60}\); ⇒ \(\frac{\frac{1}{2} + \frac{1}{\sqrt3}}{2 - \sqrt3}\); ⇒\( \frac{2 + \sqrt3}{2\sqrt3(2 - \sqrt3)}\); ⇒ \(\frac{2 + \sqrt3}{4\sqrt3- 6}\); ⇒
\(\frac{2 + \sqrt3}{4\sqrt3- 6} \times \frac{4\sqrt3 + 6}{4\sqrt3 + 6}\); ⇒\( \frac{(2 + \sqrt3)(4\sqrt3 + 6)}{(4\sqrt3)^2- 6^2} \) ; ⇒
\(\frac{8\sqrt3 + 12 + 12 + 6\sqrt3}{12} \); ⇒ \(\frac{12 + 7\sqrt3}{6} \)
SSC CGL 202032)If \(5sin\theta =4\), then the value of \({sec\theta+4cot\theta\over 4tan\theta-5cos\theta}\) is :
2
\(5 \sin \theta = 4\) ; ⇒ \( \sin \theta = 4/5\) ; ⇒ \(\frac{perpendicular}{hypotenuses} = \frac{4}{5}\);
By triplet 3-4-5,
Base = 3 ;
\(cos\theta = base/hypotenuses = 3/5\) ;
\(tan\theta = perpendicular/base = 4/3\) ;
\(\frac{\sec \theta + 4 \cot \theta}{4 \tan \theta - 5 \cos \theta}
= \frac{\frac{1}{\cos \theta} + \frac{4}{\tan \theta}}{4 \tan \theta - 5 \cos \theta}\)
\(={{{1\over{3\over5}} + {4\over{4\over3}}}\over{4\times {4\over3} - 5 \times {3\over5}}}\) = 2
SSC CGL 202033)The value of \(sec^6\theta-tan^6\theta-3sec^2\theta \space tan^2\theta+1\over cos^4\theta-sin^4\theta+2sin^2\theta+2\) is :
\(2\over3\)
\(sec^6\theta-tan^6\theta-3sec^2\theta \space tan^2\theta+1\over cos^4\theta-sin^4\theta+2sin^2\theta+2\) = \((sec^2\theta-tan^2\theta)^3+1\over(cos^2\theta-sin^2\theta)(cos^2\theta+sin^2\theta)+2sin^2\theta+2\) \([\because sec^2\theta-tan^2\theta=1] \)
\(= {1^3+1\over cos^2\theta-sin^2\theta+2sin^2\theta+2}={2\over 1+2}={2\over3}\)
SSC CGL 202034)If \(12 \cos^2 \theta - \)\(2 \sin^2 \theta + \)\(3\cos \theta = 3, \) \(0^\circ < \theta < 90^\circ\), then what is the value of \(\frac{cosec \theta + \sec \theta}{\tan \theta + \cot \theta}\) ?
\(1+\sqrt3\over2\)
\(12 \cos^2 \theta - 2 \sin^2 \theta + 3\cos \theta = 3;\)
\(12 \cos^2 \theta - 2(1 - \cos^2 \theta) + 3\cos \theta = 3;\)
\(14 \cos^2 \theta + 3\cos \theta = 5; \) Put the value of \( \theta = 60^0\),
\(14 \cos^2 60^0 + 3\cos 60^0 = 5;\)
\(14 \times \frac{1}{2} + 3 \times \frac{1}{2} = 5\);
5 = 5; ⇒ L.H.S. = R.H.S. ;
\(\frac{cosec \theta + sec \theta}{tan \theta + cot \theta} = \frac{cosec 60^0 + sec 60^0}{tan 60^0 + cot 60^0}\)
\(= \frac{\frac{2}{\sqrt3} + 2}{\sqrt3 + \frac{1}{\sqrt3}} = \frac{\frac{2 + 2\sqrt3}{\sqrt3}}{\frac{3 + 1}{\sqrt3}} = \frac{1 + \sqrt3}{2}\)