Question

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

• A. $\frac{x}{2}$
• B. $\frac{\pi }{2}-2x$
• C. $2\pi -x$
• D. $\pi -\frac{x}{2}$

Answer 1

The correct option is D $\pi -\frac{x}{2}$

We have,

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

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

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

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

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

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

$\Rightarrow {\cot }^{-1}\left(\frac{\cos x/2}{\sin x/2}\right)$

$\Rightarrow {\cot }^{-1}(-\cot x/2)$

$\Rightarrow {\cot }^{-1}\left(\cot \right(\pi -x/2\left)\right)$

$\Rightarrow \pi -x/2$.