Question

If $y={\cot }^{-1}\left(\sqrt{\cos x}\right)-{\tan }^{-1}\left(\sqrt{cosx}\right)$, prove that $\sin y={\tan }^2\frac{x}{2}$.

Answer 1

Given,

$y={\cot }^{-1}\left(\sqrt{\cos x}\right)-{\tan }^{-1}\left(\sqrt{\cos x}\right)$

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

$y=\frac{\pi }{2}-{\cos }^{-1}\left[\frac{1-\cos x}{1+\cos x}\right]$

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

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

${\tan }^2\frac{x}{2}=\cos (\frac{\pi }{2}-y)$

Therefore,

${\tan }^2\frac{x}{2}=\sin y$