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# If $\cot^{-1}(\sqrt{\cos\alpha})-\tan^{-1}(\sqrt{\cos\alpha})=x$ then $\sin x$ is equal to

$\begin{array}{1 1}(a)\;\tan^2\big(\large\frac{\alpha}{2})&(b)\;\cot^2\big(\large\frac{\alpha}{2}\big)\\(c)\;\tan\alpha&(d)\;\cot\big(\large\frac{\alpha}{2})\end{array}$

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A)
Toolbox:
• $cot^{-1}x=tan^{-1}\large\frac{1}{x}$
• $tan^{-1}x+tan^{-1}y=tan^{-1}\bigg(\large\frac{x+y}{1-xy}\bigg)$
• $1-cosx=2sin^2x/2\:\:and\:\:1+cosx=2cos^2x/2$
Given :
$\cot^{-1}(\sqrt{\cos \alpha})-\tan^{-1}(\sqrt{\cos \alpha})=x$
$\Rightarrow\:\tan^{-1}\big(\large\frac{1}{\sqrt{\cos \alpha}}\big)$$-\tan^{-1}\big(\sqrt{\cos\alpha})=x \Rightarrow\:\tan^{-1}\large\frac{1/\sqrt{\cos\alpha}-\sqrt{\cos\alpha}}{1+1/\sqrt{\cos\alpha}.\sqrt{\cos\alpha}}$$=x$
$\Rightarrow\:\large\frac{1-\cos\alpha }{2\sqrt{\cos\alpha}}$$=\tan x$
$\Rightarrow\:\sin x=\large\frac{1-\cos \alpha}{1+\cos\alpha}=\frac{2sin^2\alpha/2}{2cos^2\alpha/2}$
$\Rightarrow\:\sin x=\tan^2\large\frac{\alpha}{2}$
Hence (a) is the correct answer.