Question

If $y={\cot }^{-1}\left(\sqrt{\cos x}\right)-{\tan }^{-1}\left(\sqrt{\cos x}\right)P.T\sin y={\tan }^{2}x/2$

Answer 1

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

$\Rightarrow \cot P=\sqrt{\cos x}$

$\Rightarrow \sin P=\frac{1}{\sqrt{1+\cos x}}$

$\Rightarrow P={\sin }^{-1}\left(\frac{1}{\sqrt{1+\cos x}}\right)$

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

$\Rightarrow \tan P=\sqrt{\cos x}$

$\Rightarrow \sin Q=\frac{\sqrt{\cos x}}{\sqrt{1+\cos x}}$

$\Rightarrow P={\sin }^{-1}\left(\frac{\sqrt{\cos x}}{\sqrt{1+\cos x}}\right)$

$y=P-Q$

$\sin y=\sin (P-Q)$

$=\sin P\cos Q-\cos P\sin P$

$=\frac{1}{\sqrt{1+\cos x}}\times \frac{\sqrt{1}}{\sqrt{1+\cos x}}-\frac{\sqrt{\cos x}}{\sqrt{1+\cos x}}\times \frac{\sqrt{\cos x}}{\sqrt{1+\cos x}}$

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

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

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

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

Hence proved.