Question

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

Answer 1

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

We know, $2{\tan }^{-1}x={\tan }^{-1}\left(\frac{2x}{1-x^2}\right)$

So, $2{\tan }^{-1}\left(\sin x\right)={\tan }^{-1}\left(\frac{2\sin x}{1-si{n}^{2}x}\right)$

According to the question,

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

$\Rightarrow$ $\frac{2\sin x}{1-{\sin }^{2}x}=\frac{2}{\cos x}$

$\Rightarrow$ $\frac{\sin x}{{\cos }^{2}x}=\frac{1}{\cos x}$

$\Rightarrow$ $\sin x=\cos x$

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

$\Rightarrow$ $\tan x=1$

$\Rightarrow$ $x={\tan }^{-1}\left(1\right)$

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