cot^-1(√(cos α))-tan^-1(√(cos α))=x, then sin x is equal to:

Given that, $\cot^{-1}(\sqrt{\cos α})-\tan^{-1}(\sqrt{\cos α})=x$  … (i)

We know that, $\cot^{-1}(\sqrt{\cos α})+\tan^{-1}(\sqrt{\cos α})=\frac{π}{2}$  … (ii) $[∵ \cot^{-1}x+\tan^{-1}x=\frac{π}{2}]$

On adding equations (i) and (ii), we get : $2\cot^{-1}(\sqrt{\cos α})=\frac{π}{2}+x$

$⇒\sqrt{\cos α}=\cot(\frac{π}{4}+\frac{x}{2})⇒\sqrt{\cos α}=\frac{\cot\frac{x}{2}-1}{1+\cot\frac{x}{2}}⇒\sqrt{\cos α}=\frac{\cos\frac{x}{2}-\sin\frac{x}{2}}{\cos\frac{x}{2}+\sin\frac{x}{2}}$

On squaring both sides, we get:

$⇒\cos α=\frac{\cos^2\frac{x}{2}+\sin^2\frac{x}{2}-2\sin\frac{x}{2}\cos\frac{x}{2}}{\cos^2\frac{x}{2}+\sin^2\frac{x}{2}+2\sin\frac{x}{2}\cos\frac{x}{2}}⇒\cos α=\frac{1-\sin x}{1+\sin x}⇒\frac{1-\tan^2\frac{α}{2}}{1+\tan^2\frac{α}{2}}=\frac{1-\sin x}{1+\sin x}$

Applying componendo and dividendo rule, we get: $\sin x=\tan^2(\frac{α}{2})$