Question

The value of ${\cot }^{-1}\left\{\frac{\sqrt{1-\sin x}+\sqrt{1+\sin x}}{\sqrt{1-\sin x}-\sqrt{1+\sin x}}\right\}$, where $(0<x<\frac{\pi }{2})$ is

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

Answer 1

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

$1-sinx=(cos\frac{x}{2}-sin\frac{x}{2}{)}^2$
$1+sinx=(cos\frac{x}{2}+sin\frac{x}{2}{)}^2$
Hence,
$\frac{\sqrt{1-sinx}+\sqrt{1+sinx}}{\sqrt{1-sinx}-\sqrt{1+sinx}}$
$=\frac{(cos\frac{x}{2}-sin\frac{x}{2})+(cos\frac{x}{2}+sin\frac{x}{2})}{(cos\frac{x}{2}-sin\frac{x}{2})-(cos\frac{x}{2}+sin\frac{x}{2})}$
$=\frac{2cos\frac{x}{2}}{-2sin\frac{x}{2}}$
$=-\cot \frac{x}{2}$
The range of inverse cotangent function is $(0,\pi )$
Substitute (2) in (1)
$={cot}^{-1}(-cot\frac{x}{2})$
Since ,$\left[co{t}^{-1}\right(-cotx)=\pi -co{t}^{-1}cotx]$
$=\pi -\frac{x}{2}$