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