Question

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

Answer 1

$\begin{array}{c}co{t}^{-1}\left[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right]=\frac{x}{2}\\ \Rightarrow co{t}^{-1}\left[\frac{\left(\sqrt{1+\sin x}+\sqrt{1-\sin x}\right)\left(\sqrt{1+\sin x}+\sqrt{1-\sin x}\right)}{\left(\sqrt{1+\sin x}-\sqrt{1-\sin x}\right)\left(\sqrt{1+\sin x}+\sqrt{1-\sin x}\right)}\right]\\ \Rightarrow {\cot }^{-1}\left[\frac{1+\sin x+1-\sin x+2\sqrt{\left(1-{\sin }^{2}x\right)}}{\left(1+\sin x\right)-\left(1-\sin x\right)}\right]\\ \Rightarrow {\cot }^{-1}\left[\frac{2+2\cos x}{2\sin x}\right]\\ \Rightarrow {\cot }^{-1}\left[\frac{1+\cos x}{\sin x}\right]\\ \Rightarrow {\cot }^{-1}\left[\frac{2\cos e{c}^2\frac{x}{2}}{2\sin \frac{x}{2}\cdot \cos \frac{x}{2}}\right]\\ \Rightarrow {\cot }^{-1}\left(\cot \frac{x}{2}\right)\\ \Rightarrow \frac{x}{2}\end{array}$