Inverse Trigonometric Functions Question 11
Question: $ 2{{\tan }^{-1}}(\cos x)={{\tan }^{-1}}(cose{c^{2}}x), $ then x =
[UPSEAT 2002]
Options:
A) $ \frac{\pi }{2} $
B) $ \pi $
C) $ \frac{\pi }{6} $
D) $ \frac{\pi }{3} $
Show Answer
Answer:
Correct Answer: D
Solution:
$ 2{{\tan }^{-1}}(\cos x) $
$ ={{\tan }^{-1}}(\cos e{c^{2}}x) $
Therefore $ {{\tan }^{-1}}( \frac{2\cos x}{1-{{\cos }^{2}}x} )={{\tan }^{-1}}( \frac{1}{{{\sin }^{2}}x} ) $
$ \Rightarrow \frac{2\cos x}{{{\sin }^{2}}x}=\frac{1}{{{\sin }^{2}}x} $
Therefore $ 2\cos x=1 $ $ \Rightarrow x=\frac{\pi }{3} $ .
BETA
