Question

If $y={\cot }^{-1}\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}$, find $\frac{dy}{dx}$ if $x\in (0,\frac{\pi }{2})\cup \left(\frac{\pi }{2},\pi \right)$

Answer 1

$y={\cot }^{-1}\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}$......(1)
$\sqrt{1+\sin x}=\sqrt{{\sin }^2\frac{x}{2}+{\cos }^2\frac{x}{2}+2\sin \frac{x}{2}\cos \frac{x}{2}}$
$=\sqrt{(\sin \frac{x}{2}+\cos \frac{x}{2}{)}^2}$
$=(\sin \frac{x}{2}+\cos \frac{x}{2})$
Similarly,
$\sqrt{1-\sin x}=(\sin \frac{x}{2}-\cos \frac{x}{2})$
From(1), using above we get,
$y={\cot }^{-1}\left(\frac{2\cos \frac{x}{2}}{2\sin \frac{x}{2}}\right)$
$={\cot }^{-1}\cot \frac{x}{2}$
$=\frac{x}{2}$
$\therefore \frac{dy}{dx}=\frac{1}{2}$