Inverse Trigonometric Functions Question 18
Question: If $ 2{{\tan }^{-1}}(\cos x)={{\tan }^{-1}}(2\text{cosec }x), $ then x =
Options:
A) $ \frac{3\pi }{4} $
B) $ \frac{\pi }{4} $
C) $ \frac{\pi }{3} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
We have 2 $ {{\tan }^{-1}}(\cos x)={{\tan }^{-1}}(2\cos ecx) $
Therefore $ {{\tan }^{-1}}( \frac{2\cos x}{1-{{\cos }^{2}}x} ) $ = $ {{\tan }^{-1}}(2\text{cosec }x) $
$ \frac{2\cos x}{{{\sin }^{2}}x}=2cosecx\Rightarrow 2\cos x=2\sin x $
Or $ \sin x=\cos x $
$ \Rightarrow x=\frac{\pi }{4} $ .
BETA
