Question

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

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

Answer 1

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

$y={\tan }^{-1}\left(\frac{\cos x}{1+\sin x}\right)$
$={\tan }^{-1}\left[\frac{{\cos }^2\frac{x}{2}-{\sin }^2\frac{x}{2}}{{(\cos \frac{x}{2}+\sin \frac{x}{2})}^2}\right]$
$={\tan }^{-1}\left(\frac{\cos \frac{x}{2}-\sin \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}}\right)$
$={\tan }^{-1}\left(\frac{1-\tan \frac{x}{2}}{1+\tan \frac{x}{2}}\right)$
$={\tan }^{-1}\left[\tan (\frac{\pi }{4}-\frac{x}{2})\right]$
$\Rightarrow y=\frac{\pi }{4}-\frac{x}{2}$
On differentiating both sides w.r.t. $x$, we get
$\frac{dy}{dx}=-\frac{1}{2}$.