Question

The value of ${\cot }^{-1}\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)$ is equal to

• A. $\frac{x}{3}$
• B. $\frac{x}{4}$
• C. $1$
• D. $\frac{x}{2}$

Answer 1

Answer: (D)

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