Question

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

• A. $-\frac{1}{2}$
• B. $-1$
• C. $0$
• D. none of these

Answer 1

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

$\tan y=\frac{\cos x}{1+\sin x}$

${\sec }^{2}y\frac{dy}{dx}=\frac{-\sin x(1+\sin x)-{\cos }^{2}x}{(1+\sin x{)}^2}$

at $x=0$; $y={\tan }^{-1}\left(1\right)=\frac{\pi }{4}$

${\sec }^{2}y=2$

$\frac{dy}{dx}=\frac{0-1}{1\times 2}=\frac{-1}{2}$.