Inverse Trigonometric Functions Question 31
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} $ then $ x $ =
Options:
A) $ \sqrt{3} $
B) $ \frac{1}{\sqrt{3}} $
C) 1
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ 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} $
Putting $ x=\tan \theta $
$ 3{{\sin }^{-1}}( \frac{2\tan \theta }{1+{{\tan }^{2}}\theta } )-4{{\cos }^{-1}}( \frac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta } ) $ $ +2{{\tan }^{-1}}( \frac{2\tan \theta }{1-{{\tan }^{2}}\theta } )=\frac{\pi }{3} $
Therefore $ 3{{\sin }^{-1}}(\sin 2\theta )-4{{\cos }^{-1}}(\cos 2\theta ) $ $ +2{{\tan }^{-1}}(\tan 2\theta )=\frac{\pi }{3} $
Therefore $ 3(2\theta )-4(2\theta )+2(2\theta )=\frac{\pi }{3}\Rightarrow 6\theta -8\theta +4\theta =\frac{\pi }{3} $
Therefore $ \theta =\frac{\pi }{6}\Rightarrow {{\tan }^{-1}}x=\frac{\pi }{6}\Rightarrow x=\tan \frac{\pi }{6}=\frac{1}{\sqrt{3}} $
BETA
