Question

If $y={\tan }^{-1}\sqrt{\frac{1-\sin x}{1+\sin x}}$, then the value of $\frac{dy}{dx}atx=\frac{\pi }{6}$ is

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

Answer 1

Answer: (A)

Given, $y={\tan }^{-1}\sqrt{\frac{1-\sin x}{1+\sin x}}$

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

$={\tan }^{-1}\left|\tan \left(\frac{\pi }{4}-\frac{x}{2}\right)\right|$

$=\frac{\pi }{4}-\frac{x}{2}\left[\because x=\frac{\pi }{6}\right]$

$\Rightarrow \frac{dy}{dx}=-\frac{1}{2}$