Question

If $2{\tan }^{-1}\left(\sin x\right)={\tan }^{-1}\left(2\sec x\right),x\ne \frac{\pi }{2}$. Find the value of $x$.

Answer 1

Given: $2{\tan }^{-1}\left(\sin x\right)={\tan }^{-1}\left(2\sec x\right)$

$\Rightarrow \frac{2\sin x}{1-{\sin }^{2}x}=2\sec x\because x\ne \frac{\pi }{2}\Rightarrow 1-{\sin }^{2}x\ne 0$

$\Rightarrow \frac{2\sin x}{{\cos }^{2}x}=2\sec x$

$\Rightarrow \frac{2\sin x}{{\cos }^{2}x}=\frac{2}{\cos x}$

$\Rightarrow \frac{2\sin x}{\cos x}=2$

$\Rightarrow \frac{\sin x}{\cos x}=1$

$\Rightarrow \tan x=1$

$\Rightarrow \tan x=\tan \frac{\pi }{4}$

$\therefore x=\frac{\pi }{4}$