If $u=\cot ^{-1} \sqrt{\tan \theta}-\tan ^{-1} \sqrt{\tan \theta}$ then, $\tan \left(\frac{\pi}{4}-\frac{u}{2}\right)=$

Question.
If $u=\cot ^{-1} \sqrt{\tan \theta}-\tan ^{-1} \sqrt{\tan \theta}$ then, $\tan \left(\frac{\pi}{4}-\frac{u}{2}\right)=$
(a) $\sqrt{\tan \theta}$
(b) $\sqrt{\cot \theta}$
(c) $\tan \theta$
(d) $\cot \theta$

Solution:
(a) $\sqrt{\tan \theta}$
Let $y=\sqrt{\tan \theta}$
Then,
$u=\cot ^{-1} \sqrt{\tan \theta}-\tan ^{-1} \sqrt{\tan \theta}$
$\Rightarrow u=\cot ^{-1} y-\tan ^{-1} y$
$\Rightarrow u=\frac{\pi}{2}-2 \tan ^{-1} y \quad\left[\because \tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}\right]$
$\Rightarrow 2 \tan ^{-1} y=\frac{\pi}{2}-u$
$\Rightarrow \tan ^{-1} y=\frac{\pi}{4}-\frac{u}{2}$
$\Rightarrow y=\tan \left(\frac{\pi}{4}-\frac{u}{2}\right)$
$\Rightarrow \sqrt{\tan \theta}=\tan \left(\frac{\pi}{4}-\frac{u}{2}\right) \quad[\because y=\sqrt{\tan \theta}]$

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