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. $-\frac{1}{2}$
• E. $-1$

Answer 1

Answer: (D)

Given that, $y={\tan }^{-1}\left(\frac{\cos x}{1+\sin x}\right)$
$={\tan }^{-1}\left(\frac{{\cos }^2\frac{x}{2}-{\sin }^2\frac{x}{2}}{{\sin }^2\frac{x}{2}+{\cos }^2\frac{x}{2}+2\sin \frac{x}{2}.\cos \frac{x}{2}}\right)$
$={\tan }^{-1}\left(\frac{\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)}{{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)}^2}\right)$
$={\tan }^{-1}\left(\frac{\cos \frac{x}{2}-\sin \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}}\right)$
$={\tan }^{-1}\left(\frac{1-\tan \frac{x}{2}}{1+\tan \frac{x}{2}}\right)$
$={\tan }^{-1}\left(\tan \left(\frac{\pi }{4}-\frac{x}{2}\right)\right)$
$=\frac{\pi }{4}-\frac{x}{2}$
$\therefore$ $y=\frac{\pi }{4}-\frac{x}{2}$
Differentiating it w.r.t., $x$ we get $\frac{dy}{dx}=-\frac{1}{2}$