A) \[\sqrt{3}\]
B) \[\frac{1}{\sqrt{3}}\]
C) 1
D) None of these
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 \[x=\tan \theta \] \[3{{\sin }^{-1}}\left( \frac{2\tan \theta }{1+{{\tan }^{2}}\theta } \right)-4{{\cos }^{-1}}\left( \frac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta } \right)\]\[+2{{\tan }^{-1}}\left( \frac{2\tan \theta }{1-{{\tan }^{2}}\theta } \right)=\frac{\pi }{3}\] Þ \[3{{\sin }^{-1}}(\sin 2\theta )-4{{\cos }^{-1}}(\cos 2\theta )\]\[+2{{\tan }^{-1}}(\tan 2\theta )=\frac{\pi }{3}\] Þ \[3(2\theta )-4(2\theta )+2(2\theta )=\frac{\pi }{3}\Rightarrow 6\theta -8\theta +4\theta =\frac{\pi }{3}\] Þ \[\theta =\frac{\pi }{6}\Rightarrow {{\tan }^{-1}}x=\frac{\pi }{6}\Rightarrow x=\tan \frac{\pi }{6}=\frac{1}{\sqrt{3}}\]
You need to login to perform this action.
You will be redirected in
3 sec
