#### Explain Solution R.D.Sharma Class 12 Chapter 3 Inverse Trigonometric Function Exercise 3.14 Question 8 Sub Question 2 maths Textbook Solution.

Answer: $x=\frac{1}{\sqrt{3}}$

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

Hint: Using formula since,

$2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\sin ^{-1} \frac{2 x}{1+x^{2}}=\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)$

Solution: We have,

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

We know that  $2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\sin ^{-1} \frac{2 x}{1+x^{2}}=\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)$

\begin{aligned} &\Rightarrow 3\left(2 \tan ^{-1} x\right)-4\left(2 \tan ^{-1} x\right)+2\left(2 \tan ^{-1} x\right)=\frac{\pi}{3} \\ &\Rightarrow 6 \tan ^{-1} x-8 \tan ^{-1} x+4 \tan ^{-1} x=\frac{\pi}{3} \\ &\Rightarrow 2 \tan ^{-1} x=\frac{\pi}{3} \\ &\Rightarrow \tan ^{-1} x=\frac{\pi}{6} \end{aligned}

$\Rightarrow \tan ^{-1} x=\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)$                                            $\left[\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)=\frac{\pi}{6}\right]$

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

This is required solution.