Question

If $3\,\sin^{-1}\left(\frac{2x}{1+x^2}\right)-4\,\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)+2\,tan^{-1}\left(\frac{2x}{1-x^2}\right)=\frac{\pi}{3}$, then $x$ is equal to

• A. $\frac{1}{\sqrt3}$
• B. $-\frac{1}{\sqrt3}$
• C. $\sqrt3$
• D. $-\frac{\sqrt3}{4}$

Answer 1

Answer: (A)

Put $x=tan\,\theta $
$\therefore 3\,sin^{-1}\frac{2\,tan\,\theta}{1+tan^{2}\theta} - 4\cdot cos^{-1} \frac{1-tan^{2} \theta}{1+tan^{2}\theta}$
$+2\,tan^{-1} \frac{2\,tan\,\theta}{1-tan^{2} \theta} = \frac{\pi}{3}$
$ \Rightarrow 3\,sin^{-1} \left(sin\,2\,\theta\right) -4\,cos^{-1}\left(cos\,2\,\theta\right)+2\,sin^{-1} \left(tan\, 2\,\theta\right) = \frac{\pi}{3} $
$ \Rightarrow 3\left(2\,\theta\right)-4\left(2\,\theta\right)+2\left(2\,\theta\right) = \frac{\pi}{3} $
$\Rightarrow 2\,\theta= \frac{\pi}{3} $
$ \Rightarrow \theta = \frac{\pi}{6} $
$ \Rightarrow x=tan \frac{\pi}{6} $
$= \frac{1}{\sqrt{3}}$