Question

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

• A. $-\frac{1}{2}$
• B. $-\frac{1}{4}$
• C. $-\frac{1}{8}$
• D. $\frac{1}{2}$

Answer 1

Answer: (A)

$-\frac{1}{2}$

Let $y={\tan }^{-1}\left(\frac{\cos x}{1+\sin x}\right)$

$\Rightarrow y={\tan }^{-1}\left(\frac{({\cos }^2\frac{x}{2}-{\sin }^2\frac{x}{2})}{{\sin }^2\frac{x}{2}+{\cos }^2\frac{x}{2}+2\sin \frac{x}{2}.\cos \frac{x}{2}}\right)$

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

$={\tan }^{-1}\left(\frac{1-\tan \left(x/2\right)}{1+\tan \left(x/2\right)}\right)$

$={\tan }^{-1}\left(\tan (\pi /4-x/2)\right)=\pi /4-x/2$

$\therefore \frac{dy}{dx}=-\frac{1}{2}$