(b)
We have, 2 $\tan^{- 1}{(\cos{x) = \tan^{- 1}{(2\ cosec\ x)}}}$
$\Rightarrow \tan^{- 1}{\left( \frac{2\cos x}{1 - \cos^{2}x} \right) = \tan^{- 1}{(2\ cosec\ x)}}$
$\Rightarrow \frac{2\cos x}{\sin^{2}x} = 2\ cosec\ x$
$\Rightarrow \sin{x = \cos{x \Rightarrow x = \frac{\pi}{4}}}$