Question

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

Answer 1

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

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

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

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

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

${\tan }^{-1}\left(\frac{\tan \left(\frac{\pi }{4}\right)-\tan \left(\frac{x}{2}\right)}{1+\tan \left(\frac{\pi }{4}\right)\tan \frac{x}{2}}\right)$

${\tan }^{-1}(\tan \left(\frac{\pi }{4}\right)-\tan \left(\frac{x}{2}\right))=\frac{\pi }{4}-\frac{x}{2}$