$\text{ If }{\cot }^{-1}\left(\sqrt{\mathrm{cos\alpha }}\right)-{\tan }^{-1}\left(\sqrt{\mathrm{cos\alpha }}\right)=\mathrm{x}\ge 0\text{ then }\mathrm{sinx}=$

Given that ${\cot }^{-1}\left(\sqrt{\mathrm{cos\alpha }}\right)-{\tan }^{-1}\left(\sqrt{\mathrm{cos\alpha }}\right)=\mathrm{x}$

We know that ${\tan }^{-1}A+co{t}^{-1}A=\frac{\mathrm{\pi }}{2}$

Hence, the given equaiton can be rewrite as 

$\frac{\mathrm{\pi }}{2}-2{\tan }^{-1}\left(\sqrt{\mathrm{cos\alpha }}\right)=\mathrm{x}$

It implies that 

$\begin{array}{rcl}{\tan }^{-1}\left(\sqrt{\mathrm{cos\alpha }}\right)& =& \frac{\mathrm{\pi }}{4}-\frac{x}{2}\\ \sqrt{\cos \alpha }& =& \frac{1-\tan \left({\displaystyle \frac{x}{2}}\right)}{1+\tan \left({\displaystyle \frac{x}{2}}\right)}\\ & =& \frac{\cos \left(\frac{x}{2}\right)-\sin \left(\frac{x}{2}\right)}{\cos \left(\frac{x}{2}\right)+\sin \left(\frac{x}{2}\right)}\end{array}$

Squaring on both sides

$\cos \alpha =\frac{1-\sin x}{1+\sin x}$

Use componendo and dividendo

$\begin{array}{rcl}\frac{1-\sin x+1+\sin x}{1-\sin x-1-\sin x}& =& \frac{\cos \alpha +1}{\cos \alpha -1}\\ \sin x& =& \frac{2{\sin }^2\left({\displaystyle \frac{\alpha }{2}}\right)}{2{\cos }^2\left(\frac{\alpha }{2}\right)}\\ & =& {\tan }^2\left(\frac{\alpha }{2}\right)\end{array}$