$2{\tan ^{ - 1}}(\cos x) = {\tan ^{ - 1}}(2{\rm{cosec }}x)$

We have, $2{\tan ^{ - 1}}(\cos x) = {\tan ^{ - 1}}(2{\rm{cosec }}x)$
$\Rightarrow$ ${\tan ^{ - 1}}\left[ {\frac{{2\cos x}}{{1 - {{\cos }^2}x}}} \right] = {\tan ^{ - 1}}(2{\rm{cosec }}x)$

$\Rightarrow$ $\tan \left[ {{{\tan }^{ - 1}}\left( {\frac{{2\cos x}}{{{{\sin }^2}x}}} \right)} \right] = 2\,\,{\rm{cosec }}x$

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

$\Rightarrow$ $\tan x = 1\;\; \Rightarrow x = {\tan ^{ - 1}}1 \Rightarrow x = \frac{\pi }{4}$