Question

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

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

Answer 1

Answer: (A)

$\pi -\frac{x}{2}$

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

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

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

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

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

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

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

$={\cot }^{-1}(-\cot \frac{x}{2})$

$=\pi -\frac{x}{2}$

Hence, the correct answer is $\pi -\frac{x}{2}$.