Question

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

Answer 1

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

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