Question

Find the value of $\theta$ in each of the following if
(i) $2\cos 3\theta =1$

(ii) $2\sqrt{3}\tan \theta =6$
(iii) $\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }=\frac{1}{2}$

• A. (i) ${10}^{\circ }$ (ii) ${40}^{\circ }$ (iii) ${70}^{\circ }$
• B. (i) ${35}^{\circ }$ (ii) ${55}^{\circ }$ (iii) ${80}^{\circ }$
• C. (i) ${20}^{\circ }$ (ii) ${60}^{\circ }$ (iii) ${30}^{\circ }$
• D. (i) ${40}^{\circ }$ (ii) ${70}^{\circ }$ (iii) ${100}^{\circ }$

Answer 1

The correct option is C (i) ${20}^{\circ }$ (ii) ${60}^{\circ }$ (iii) ${30}^{\circ }$

(i) $2\cos 3\theta =1$
$\Rightarrow \cos 3\theta =\frac{1}{2}$
$\Rightarrow \cos 3\theta =\cos 60$
Therefore, $3\theta =60$
$\Rightarrow \theta =20$
(ii) $2\sqrt{3}\tan \theta =6$
$\Rightarrow \tan \theta =\frac{6}{2\sqrt{3}}$
$\Rightarrow \tan \theta =\frac{3\times \sqrt{3}}{\sqrt{3}\times \sqrt{3}}$
$\Rightarrow \tan \theta =\sqrt{3}$
(iii) $\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }=\frac{1}{2}$
$\Rightarrow \frac{1-{\tan }^2\theta }{{\sec }^2\theta }=\frac{1}{2}$
$\Rightarrow \frac{1-\frac{{\sin }^2\theta }{{\cos }^2\theta }}{{\sec }^2\theta }=\frac{1}{2}$
$\Rightarrow \frac{{\cos }^2\theta -{\sin }^2\theta }{{\cos }^2\theta \times {\sec }^2\theta }=\frac{1}{2}$
$\Rightarrow \frac{{\cos }^2\theta -{\sin }^2\theta }{\frac{1}{{\sec }^2\theta }\times {\sec }^2\theta }=\frac{1}{2}$
$\Rightarrow {\cos }^2\theta -{\sin }^2\theta =\frac{1}{2}$
$\Rightarrow {\cos }^2\theta -1+{\cos }^2\theta =\frac{1}{2}$
$\Rightarrow 2{\cos }^2\theta =\frac{1}{2}+1$
$\Rightarrow 2{\cos }^2\theta =\frac{3}{2}$
$\Rightarrow {\cos }^2\theta =\frac{3}{4}$
$\Rightarrow \cos \theta =\sqrt{\frac{3}{4}}$
$\Rightarrow \cos \theta =\frac{\sqrt{3}}{2}$ .....$[$Since $\frac{\sqrt{3}}{2}]$
$\Rightarrow \theta =30$