Question

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

Answer 1

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

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

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

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

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

Now apply, $\cos 2A=\frac{1-{\tan }^{2}A}{1+{\tan }^{2}A}$ on R.H.S.

$\sin y=\frac{1-{\tan }^2\left[{\tan }^{-1}\left(\sqrt{\cos x}\right)\right]}{1+{\tan }^2\left[{\tan }^{-1}\left(\sqrt{\cos x}\right)\right]}$

$\sin y=\frac{1-cosx}{1+cosx}$

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

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