Question
Solve the following equation:
$2\tan^{-1}\left(\cos x\right)=\tan^{-1}\left(2\ \text{cosec}\ x\right)$

Answer

$2\tan^{-1}\left(\cos x\right)=\tan^{-1}\left(2\ \text{cosec}\ x\right)$
$\Rightarrow\tan^{-1}\bigg(\frac{2\cos x}{1-\cos^2 x}\bigg)=\tan^{-1}\left(2\ \text{cosec}\ x\right)$ $\left[2\tan^{-1}x=\tan^{-1}\frac{2x}{1-x^2}\right]$
$\Rightarrow\frac{2\cos x}{1-\cos^2x}=2\ {\text{cosec}\ x}$
$\Rightarrow\frac{2\cos x}{\sin^2x}=\frac{2}{\sin x}$
$\Rightarrow\cos x=\sin x$
$\Rightarrow\tan x=1$
$\therefore x=\frac{\pi}{4}$

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