# if the

Question:

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

(a) $\frac{1}{\sqrt{3}}$

(b) $-\frac{1}{\sqrt{3}}$

(c) $\sqrt{3}$

(d) $-\frac{\sqrt{3}}{4}$

Solution:

(a) $\frac{1}{\sqrt{3}}$

Let $x=\tan y$

Then,

$3 \sin ^{-1}\left(\frac{2 \tan y}{1+\tan ^{2} y}\right)-4\left(\frac{1-\tan ^{2} y}{1+\tan ^{2} y}\right)+2 \tan ^{-1}\left(\frac{2 \tan y}{1-\tan ^{2} y}\right)=\frac{\pi}{3}$

$\Rightarrow 3 \sin ^{-1}(\sin 2 y)-4 \cos ^{-1}(\cos 2 y)+2 \tan ^{-1}(\tan 2 y)=\frac{\pi}{3}$

$\left[\because \sin 2 y=\left(\frac{2 \tan y}{1+\tan ^{2} y}\right), \cos 2 y=\left(\frac{1-\tan ^{2} y}{1+\tan ^{2} y}\right)\right.$ and $\left.\tan 2 y=\left(\frac{2 \tan y}{1-\tan ^{2} y}\right)\right]$

$\Rightarrow 3 \times 2 y-4 \times 2 y+2 \times 2 y=\frac{\pi}{3}$

$\Rightarrow 6 y-8 y+4 y=\frac{\pi}{3}$

$\Rightarrow 2 y=\frac{\pi}{3}$

$\Rightarrow y=\frac{\pi}{6}$

$\Rightarrow \tan ^{-1} x=\frac{\pi}{6}$           $\left[\because \tan ^{-1} x=y\right]$

$\Rightarrow x=\tan \frac{\pi}{6}$

$\Rightarrow x=\frac{1}{\sqrt{3}}$