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\\}\left(0 < x < \frac{\pi}{2}\right) is

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

Answer 1

Answer: (A)

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

Answer 2

Since, $1\pm sin\,x=\left(cos \frac{x}{2}\pm sin \frac{x}{2}\right)^{2}$ \therefore cot^{-1}\left\\{\frac{\sqrt{1-sin\,x}+\sqrt{1+sin\,x}}{\sqrt{1-sin\,x}-\sqrt{1+sin\,x}}\right\\} $=cot^{-1}\left[\frac{\left(cos \frac{x}{2}-sin \frac{x}{2}\right)+\left(cos \frac{x}{2}+sin \frac{x}{2}\right)}{\left(cos \frac{x}{2}-sin \frac{x}{2}\right)-\left(cos \frac{x}{2}+sin \frac{x}{2}\right)}\right]$ =cot^{-1}\left\\{-cot \frac{x}{2}\right\\}=cot^{-1}\left\\{cot\left(\pi-\frac{\pi}{2}\right)\right\\}=\pi-\frac{x}{2}