Question

If $\frac{1}{2}{\sin }^{-1}\left[\frac{3\sin 2\theta }{5+4\cos 2\theta }\right]=ta{n}^{-1}x,$ then $x=$

• A. $\tan 3\theta$
• B. $3\tan \theta$
• C. $\left(1/3\right)\tan \theta$
• D. $3\cot \theta$

Answer 1

The correct option is C $\left(1/3\right)\tan \theta$

Given, $\frac{1}{2}{\sin }^{-1}\left[\frac{3\sin 2\theta }{5+4\cos 2\theta }\right]=ta{n}^{-1}x,$

Let $\tan \theta =3\tan \varphi$ ..(1)
$\therefore \frac{3\sin 2\theta }{5+4\cos 2\theta }=\frac{6\tan \theta }{9+{\tan }^2\theta }=\frac{2\tan \varphi }{1+{\tan }^2\varphi }=\sin 2\varphi$

Hence
$\frac{1}{2}{\sin }^{-1}\left[\frac{3\sin 2\theta }{5+4\cos 2\theta }\right]=\frac{1}{2}{\sin }^{-1}\sin 2\varphi =\varphi$

And from (1)
$\varphi ={\tan }^{-1}\left(\frac{1}{3}\tan \theta \right)$
Therefore,
$x=\frac{1}{3}\tan \theta$