Question

For $-\frac{\pi }{2}<x<\frac{3\pi }{2}$, the value of $\frac{d}{dx}\left\{{\tan }^{-1}\frac{\cos x}{1+\sin x}\right\}$ is equal to

• A. $\frac{1}{2}$
• B. $-\frac{1}{2}$
• C. $1$
• D. $\frac{\sin x}{(1+\sin x{)}^2}$

Answer 1

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

$\frac{d}{dx}\left\{{\tan }^{-1}\frac{\cos x}{1+\sin x}\right\}$

$=\frac{1}{1+{\left(\frac{\cos x}{1+\sin x}\right)}^2}\frac{d}{dx}\frac{\cos x}{1+\sin x}$

$=\frac{(1+\sin x{)}^2}{1+{\sin }^{2}x+2\sin x+{\cos }^{2}x}$$\times \frac{(1+\sin x)(-\sin x)-\cos x\left(\cos x\right)}{(1+\sin x{)}^2}$

$=\frac{1}{2+2\sin x}\times [-(\sin x+{\sin }^{2}x+{\cos }^{2}x\left)\right]$

$=\frac{1}{2(1+\sin x)}\times -(1+\sin x)=-\frac{1}{2}$