Question

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

Answer 1

Consider $\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}$

$=\frac{{(\sqrt{1+\sin x}+\sqrt{1-\sin x})}^2}{(\sqrt{1+\sin x}{)}^2-(\sqrt{1-\sin x}{)}^2}$ ...(by rationalizing)

$=\frac{(1+\sin x)+(1-\sin x)+2\sqrt{(1+\sin x)(1-\sin x)}}{1+\sin x-1+\sin x}$

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

$=\cot \frac{x}{2}$

$\therefore \text{L.H.S.}={\cot }^{-1}\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)={\cot }^{-1}\left(\cot \frac{x}{2}\right)=\frac{x}{2}=\text{R.H.S.}$