Question

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

• A. $\frac{1}{\sqrt{3}}$
• B. $-\frac{1}{\sqrt{3}}$
• C. $\sqrt{3}$
• D. $-\frac{\sqrt{3}}{4}$

Answer 1

Answer: (A)

Put $x=tan\theta$
$\therefore 3si{n}^{-1}\frac{2tan\theta }{1+ta{n}^2\theta }-4\cdot co{s}^{-1}\frac{1-ta{n}^2\theta }{1+ta{n}^2\theta }$
$+2ta{n}^{-1}\frac{2tan\theta }{1-ta{n}^2\theta }=\frac{\pi }{3}$
$\Rightarrow 3si{n}^{-1}\left(sin2\theta \right)-4co{s}^{-1}\left(cos2\theta \right)+2si{n}^{-1}\left(tan2\theta \right)=\frac{\pi }{3}$
$\Rightarrow 3\left(2\theta \right)-4\left(2\theta \right)+2\left(2\theta \right)=\frac{\pi }{3}$
$\Rightarrow 2\theta =\frac{\pi }{3}$
$\Rightarrow \theta =\frac{\pi }{6}$
$\Rightarrow x=tan\frac{\pi }{6}$
$=\frac{1}{\sqrt{3}}$