Question

Solve ${\tan }^{-1}\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)$

Answer 1

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

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

$={\tan }^{-1}\frac{\sqrt{{(\sin \frac{x}{2}+\cos \frac{x}{2})}^2}+\sqrt{{(\sin \frac{x}{2}-\cos \frac{x}{2})}^2}}{\sqrt{{(\sin \frac{x}{2}+\cos \frac{x}{2})}^2}-\sqrt{{(\sin \frac{x}{2}-\cos \frac{x}{2})}^2}}$

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

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

$={\tan }^{-1}\tan \frac{x}{2}$

$=\frac{x}{2}$

Hence, it is complete solution.