Question

If $3si{n}^{-1}\frac{2x}{1+x^2}-4co{s}^{-1}\frac{1-x^2}{1+x^2}+2ta{n}^{-1}\frac{2x}{1+x^2}=\frac{\pi }{3}$ then x=

• A. $\sqrt{3}$
• B. $\frac{1}{\sqrt{3}}$
• C. 1
• D. None of these

Answer 1

The correct option is B $\frac{1}{\sqrt{3}}$

$3si{n}^{-1}\frac{2x}{1+x^2}-4co{s}^{-1}\frac{1-x^2}{1+x^2}+2ta{n}^{-1}\frac{2x}{1+x^2}=\frac{\pi }{3}$
Putting $x=tan\theta$
$3si{n}^{-1}\left(\frac{2tan\theta }{1+ta{n}^2\theta }\right)-4co{s}^{-1}\left(\frac{1-ta{n}^2\theta }{1+ta{n}^2\theta }\right)$
$+2ta{n}^{-1}\left(\frac{2tan\theta }{1-ta{n}^2\theta }\right)=\frac{\pi }{3}$
$\Rightarrow 3si{n}^{-1}\left(sin2\theta \right)-4co{s}^{-1}\left(cos2\theta \right)$
$+2ta{n}^{-1}\left(tan2\theta \right)=\frac{\pi }{3}$
$\Rightarrow 3\left(2\theta \right)-4\left(2\theta \right)+2\left(2\theta \right)=\frac{\pi }{3}\Rightarrow 6\theta -8\theta +4\theta =\frac{\pi }{3}$
$\Rightarrow \theta =\frac{\pi }{6}\Rightarrow ta{n}^{-1}x=\frac{\pi }{6}\Rightarrow x=tan\frac{\pi }{6}=\frac{1}{\sqrt{3}}$