Question

Differentiate $y={\tan }^{-1}\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)$ with respect $x$.
$\left(1\right)$ When $0<x<\frac{\pi }{2}$
$\left(2\right)$ When $\frac{\pi }{2}<x<\pi$

Answer 1

We know that,
$\sqrt{1+\sin x}=\sqrt{{\sin }^2\frac{x}{2}+{\cos }^2\frac{x}{2}+2\sin \frac{x}{2}\cos \frac{x}{2}}$
$=\sqrt{{(\cos \frac{x}{2}+\sin \frac{x}{2})}^2}$
$=|\cos \frac{x}{2}+\sin \frac{x}{2}|$
$=\cos \frac{x}{2}+\sin \frac{x}{2}$ for $0<x<\pi$

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

$\sqrt{1-\sin x}=\{\begin{array}{cc}\cos \frac{x}{2}-\sin \frac{x}{2},& \text{if}0<x<\frac{\pi }{2}\\ \sin \frac{x}{2}-\cos \frac{x}{2},& \text{if}\frac{\pi }{2}<x<\pi \end{array}$

Thus,
Case 1: when $0<x<\frac{\pi }{2}$
$y={\tan }^{-1}\left[\frac{(\cos \frac{x}{2}+\sin \frac{x}{2})+(\cos \frac{x}{2}-\sin \frac{x}{2})}{(\cos \frac{x}{2}+\sin \frac{x}{2})-(\cos \frac{x}{2}-\sin \frac{x}{2})}\right]$
$={\tan }^{-1}\left(\cot \frac{x}{2}\right)$
$={\tan }^{-1}\left[\tan (\frac{\pi }{2}-\frac{x}{2})\right]$
$=\frac{\pi }{2}-\frac{x}{2}$
$\Rightarrow \frac{dy}{dx}=-\frac{1}{2}$

Case 2: when $\frac{\pi }{2}<x<\pi$
$y={\tan }^{-1}\left[\frac{(\cos \frac{x}{2}+\sin \frac{x}{2})+(\sin \frac{x}{2}-\cos \frac{x}{2})}{(\cos \frac{x}{2}+\sin \frac{x}{2})-(\sin \frac{x}{2}-\cos \frac{x}{2})}\right]$
$={\tan }^{-1}\left(\tan \frac{x}{2}\right)$
$=\frac{x}{2}$
$\Rightarrow \frac{dy}{dx}=\frac{1}{2}$