Question

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

Answer 1

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

$taking\sin bothside$

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

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

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

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

$\sin \left[{\sin }^{-1}\left(\frac{1-\cos x}{1+\cos x}\right)\right]$

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

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

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