Question

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

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

Answer 1

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

Given, $y={\tan }^{-1}\sqrt{\frac{1-\sin x}{1+\sin x}}$
$={\tan }^{-1}\sqrt{\frac{1-\cos (\frac{\pi }{2}-x)}{1+\cos (\frac{\pi }{2}-x)}}$
$={\tan }^{-1}\left|\tan (\frac{\pi }{4}-\frac{x}{2})\right|$
$=\frac{\pi }{4}-\frac{x}{2}$ $[\because x=\frac{\pi }{6}]$
$\Rightarrow \frac{dy}{dx}=-\frac{1}{2}$