If $2 \tan ^{-1}(\cos x)=\tan ^{-1}(2 \operatorname{cosec} x)$, then the value of $x$ is

Given the equation: $2\tan^{-1}(\cos x) = \tan^{-1}(2\csc x)$

Using the identity $2\tan^{-1}(y) = \tan^{-1}\left(\frac{2y}{1-y^2}\right)$ for $y^2 < 1$, let $y = \cos x$ (so $\cos^2 x < 1$, implying $\cos x \neq \pm1$):
$2\tan^{-1}(\cos x) = \tan^{-1}\left(\frac{2\cos x}{1-\cos^2 x}\right)$

Since $1 - \cos^2 x = \sin^2 x$, we have
$\tan^{-1}\left(\frac{2\cos x}{\sin^2 x}\right) = \tan^{-1}(2\csc x)$

Equating arguments (as $\tan^{-1}$ is injective):
$\frac{2\cos x}{\sin^2 x} = 2\csc x$

Substitute $\csc x = \frac{1}{\sin x}$ (with $\sin x \neq 0$):
$\frac{2\cos x}{\sin^2 x} = \frac{2}{\sin x}$
Multiplying both sides by $\sin^2 x$:
$2\cos x = 2\sin x$
Thus, $\cos x = \sin x$

This implies $\tan x = 1$, so $x = n\pi + \frac{\pi}{4}$ for all integers $n$.

Verifying $\cos^2 x < 1$ and $\sin x \neq 0$:
$\cos^2(\pi/4) = 1/2 < 1$ and $\sin(\pi/4) = 1/\sqrt{2} \neq 0$, confirming validity.

Solution: $x = \frac{\pi}{4}$