Question

Prove that
${\tan }^{-1}\left(\frac{\cos x}{1+\sin x}\right)=\frac{\pi }{4}-\frac{x}{2}$.

Answer 1

${\tan }^{-1}\left(\frac{\cos x}{1+\sin x}\right)={\tan }^{-1}\left(\frac{{\cos }^2\frac{x}{2}-{\sin }^2\frac{x}{2}}{{\sin }^2\frac{x}{2}+{\cos }^2\frac{x}{2}+2\sin \frac{x}{2}\cos \frac{x}{2}}\right)={\tan }^{-1}\left[\frac{(\cos \frac{x}{2}-\sin \frac{x}{2})(\cos \frac{x}{2}+\sin \frac{x}{2})}{{(\cos \frac{x}{2}+\sin \frac{x}{2})}^2}\right]={\tan }^{-1}\left[\frac{\cos \frac{x}{2}-\sin \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}}\right]={\tan }^{-1}\left[\frac{1-\tan \frac{x}{2}}{1+\tan \frac{x}{2}}\right]={\tan }^{-1}\left[\frac{\tan \frac{\pi }{4}-\tan \frac{x}{4}}{1+\tan \frac{\pi }{4}\tan \frac{x}{2}}\right]={\tan }^{-1}\left[\tan (\frac{\pi }{4}-\frac{x}{2})\right]=\frac{\pi }{4}-\frac{x}{2}$