Question

$2$ tan${}^{-1}$ $\left(\cos x\right)={\tan }^{-1}\left(2\cos ecx\right)$

Answer 1

$2{\tan }^{-1}\left(\cos x\right)={\tan }^{-1}\left(2cosecx\right)$

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

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

$\therefore 2{\tan }^{-1}\left(\cos x\right)={\tan }^{-1}\left(\frac{2\cos x}{{\sin }^{2}x}\right)$ ……..$\left(1\right)$

Given

$2{\tan }^{-1}\left(\cos x\right)={\tan }^{-1}\left(2cosecx\right)$

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

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

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

$\cos x=\sin x$

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