Question

If $y={\tan }^{-1}\left(\frac{\cos x}{1+\sin x}\right),x\in (-\frac{\pi }{2},\frac{\pi }{2})$, then $\frac{dy}{dx}$ is equal to

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

Answer 1

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

Convert $\frac{\cos x}{1+\sin x}$ in terms of $\tan$ and proceed.
$\because$$\frac{\cos x}{1+\sin x}=\frac{\sin (\frac{\pi }{2}-x)}{1+\cos (\frac{\pi }{2}-x)}$
$=\tan (\frac{\pi }{4}-\frac{x}{2})$
$\therefore {\tan }^{-1}\left(\frac{\cos x}{1+\sin x}\right)={\tan }^{-1}\left(\tan (\frac{\pi }{4}-\frac{x}{2})\right)$
$\because x\in (-\frac{\pi }{2},\frac{\pi }{2})\therefore \frac{\pi }{4}-\frac{x}{2}\in (0,\frac{\pi }{2})$
$\Rightarrow$ $y=(\frac{\pi }{4}-\frac{x}{2})$
$\Rightarrow$ $\frac{dy}{dx}=-\frac{1}{2}$