A) \[-\frac{1}{2}\]
B) \[\frac{1}{2}\]
C) \[-1\]
D) 1
Correct Answer: A
Solution :
\[\frac{d}{dx}\left[ {{\tan }^{-1}}\left( \frac{\cos x}{1+\sin x} \right) \right]\] \[=\frac{d}{dx}\left[ {{\tan }^{-1}}\left( \frac{{{\cos }^{2}}\frac{x}{2}-{{\sin }^{2}}\frac{x}{2}}{{{\cos }^{2}}\frac{x}{2}+{{\sin }^{2}}\frac{x}{2}+2\sin \frac{x}{2}\cos \frac{x}{2}} \right) \right]\] \[=\frac{d}{dx}\left[ {{\tan }^{-1}}\left( \frac{1-\tan \left( \frac{x}{2} \right)}{1+\tan \left( \frac{x}{2} \right)} \right) \right]=\frac{d}{dx}\left[ {{\tan }^{-1}}\tan \left( \frac{\pi }{4}-\frac{x}{2} \right) \right]=-\frac{1}{2}\]
You need to login to perform this action.
You will be redirected in
3 sec
