Question

Prove that ${\cot }^{-1}\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)=\frac{x}{2};x\in (0,\frac{\pi }{4}).$

Answer 1

LHS $=co{t}^{-1}\left[\frac{\sqrt{1+sinx}+\sqrt{1-sinx}}{\sqrt{1+sinx}-\sqrt{1-sinx}}\right],x\in (0,\frac{\pi }{4})$

Given, $0<x<\frac{\pi }{4}$

$\Rightarrow 0<x<\frac{\pi }{8}$

$\Rightarrow \frac{x}{2}\in (0,\frac{\pi }{4})\subset (0,\pi )$

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

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

$=co{t}^{-1}\left(\frac{cos\frac{x}{2}}{sin\frac{x}{2}}\right)\Rightarrow co{t}^{-1}\left(cot\frac{x}{2}\right)\Rightarrow \frac{x}{2}=$ RHS

Hence proved.