Inverse Trigonometric Functions Question 156
Question: $ {{\tan }^{-1}}[ \frac{\cos x}{1+\sin x} ]= $
Options:
A) $ \frac{\pi }{4}-\frac{x}{2} $
B) $ \frac{\pi }{4}+\frac{x}{2} $
C) $ \frac{x}{2} $
D) $ \frac{\pi }{4}-x $
Show Answer
Answer:
Correct Answer: A
Solution:
$ {{\tan }^{-1}}[ \frac{\cos x}{1+\sin x} ]={{\tan }^{-1}}[ \frac{\sin (\pi /2-x)}{1+\cos (\pi /2-x)} ] $
$ ={{\tan }^{-1}}[ \frac{2\sin (\pi /4-x/2)\cos (\pi /4-x/2)}{2{{\cos }^{2}}(\pi /4-x/2)} ] $
$ ={{\tan }^{-1}}\tan ( \frac{\pi }{4}-\frac{x}{2} )=\frac{\pi }{4}-\frac{x}{2} $ .
BETA
