$2 \tan ^{-1}\left\{\operatorname{cosec}\left(\tan ^{-1} x\right)-\tan \left(\cot ^{-1} x\right)\right\}$ is equal to

Question.
$2 \tan ^{-1}\left\{\operatorname{cosec}\left(\tan ^{-1} x\right)-\tan \left(\cot ^{-1} x\right)\right\}$ is equal to
(a) $\cot ^{-1} x$
(b) $\cot ^{-1} \frac{1}{x}$
(c) $\tan ^{-1} x$
(d) none of these
Solution:
(c) $\tan ^{-1} x$
Let $\tan ^{-1} x=y$
So, $x=\tan y$
$\therefore 2 \tan ^{-1}\left\{\operatorname{cosec}\left(\tan ^{-1} x\right)-\tan \left(\cot ^{-1} x\right)\right\}=2 \tan ^{-1}\left\{\operatorname{cosec}\left(\tan ^{-1} x\right)-\tan \left(\tan ^{-1} \frac{1}{x}\right)\right\}$
$=2 \tan ^{-1}\left\{\operatorname{cosec}\left(\tan ^{-1} x\right)-\frac{1}{x}\right\}$
$=2 \tan ^{-1}\left\{\operatorname{cosec} y-\frac{1}{\tan y}\right\}$
$=2 \tan ^{-1}\left\{\frac{1-\cos y}{\sin y}\right\}$
$=2 \tan ^{-1}\left\{\frac{2 \sin ^{2} \frac{y}{2}}{\sin y}\right\}$
$=2 \tan ^{-1}\left\{\frac{2 \sin ^{2} \frac{y}{2}}{2 \sin \frac{y}{2} \cos \frac{y}{2}}\right\}$
$=2 \tan ^{-1}\left\{\tan \frac{y}{2}\right\}$
$=y$
$=\tan ^{-1} x$

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