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

Answer: (A)

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

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

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

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

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