Inverse Trigonometric Functions Question 128
Question: If $ 3{{\sin }^{-1}}( \frac{2x}{1+x^{2}} )-4{{\cos }^{-1}}( \frac{1-x^{2}}{1+x^{2}} )+2{{\tan }^{-1}}( \frac{2x}{1-x^{2}} )=\frac{\pi }{3} $ where $ | x |<1, $ then x is equal to
Options:
A) $ \frac{1}{\sqrt{3}} $
B) $ -\frac{1}{\sqrt{3}} $
C) $ \sqrt{3} $
D) $ -\frac{\sqrt{3}}{4} $
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Answer:
Correct Answer: A
The given equation is $ 3(2\tan^{-1}x)-4(2\tan^{-1}x)+2(\tan^{-1}x)=\pi /3 $
$ \therefore 2{{\tan }^{-1}}x=\pi /3 $
$ \therefore {{\tan }^{-1}}x=\pi /6 $
$ \therefore x=\frac{1}{\sqrt{3}} $
BETA
