Question

If $y\left(x\right)={\cot }^{-1}\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right),x\in (\frac{\pi }{2},\pi ),$ then $\frac{dy}{dx}$ at $x=\frac{5\pi }{6}$ is

• A. $-\frac{1}{2}$
• B. $0$
• C. $-1$
• D. $\frac{1}{2}$

Answer 1

The correct option is A $-\frac{1}{2}$

$\because \sqrt{1+\sin x}=\sin \frac{x}{2}+\cos \frac{x}{2}$ and $\sqrt{1-\sin x}=\sin \frac{x}{2}-\cos \frac{x}{2}$

$y\left(x\right)={\cot }^{-1}\left(\frac{2\sin \frac{x}{2}}{2\cos \frac{x}{2}}\right)={\cot }^{-1}\left(\tan \frac{x}{2}\right)=\frac{\pi }{2}-\frac{x}{2}$
$\frac{dy}{dx}=-\frac{1}{2}$