Question

Simplify

${\cot }^{-1}\left[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right]=\frac{x}{m};x\in (0,\frac{\pi }{4})$.Find $m$

Answer 1

${\cot }^{-1}\left[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right]$

$={\cot }^{-1}\left[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\times \frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}+\sqrt{1-\sin x}}\right]$

$={\cot }^{-1}\left[\frac{1+\sin x+1-\sin x+2\sqrt{1+\sin x}\sqrt{1-\sin x}}{1+\sin x-\left(1-\sin x\right)}\right]$

$={\cot }^{-1}\left[\frac{2+2\cos x}{2\sin x}\right]$

$={\cot }^{-1}\left[\frac{1+\cos x}{\sin x}\right]$

$={\cot }^{-1}\left[\frac{2{\cos }^2\frac{x}{2}}{2\sin \frac{x}{2}\cos \frac{x}{2}}\right]$

$={\cot }^{-1}\left(\cot \frac{x}{2}\right)$

$=\frac{x}{2}$

on comparing it with $=\frac{x}{m}$ we get $m=2$