Question

If ${\cot }^{-1}\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)=\frac{x}{m},x\in [0,\frac{\pi }{2}]$. Find $m$.

Answer 1

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

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

$\Rightarrow =\frac{1+\sin x+1-\sin x+2\sqrt{1-{\sin }^{2}x}}{1+\sin x-1+\sin x}$

$\Rightarrow =\frac{2+2\cos x}{2\sin x}$

$\Rightarrow =\frac{1+\cos x}{\sin x}$

$\Rightarrow =\frac{2{\cos }^2\frac{x}{2}}{2\sin \frac{x}{2}\cos \frac{x}{2}}$

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

$\therefore {\cot }^{-1}\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)={\cot }^{-1}\left(\cot \frac{x}{2}\right)=\frac{x}{2},\text{For all}x\in [0,\frac{\pi }{2}]$

$\therefore m=2$