#### Explain Solution R.D.Sharma Class 12 Chapter 3 Inverse Trigonometric Function Exercise 3.14 Question 8 Sub Question 5 maths Textbook Solution.

Answer: $x=\sqrt{3}$

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

Hint: Using formula

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

$\text { 2. } \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=2 \tan ^{-1} x$

Solution:

We have,

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

We know that,

\begin{aligned} &\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)=2 \tan ^{-1} x \\ &\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=2 \tan ^{-1} x \\ &\Rightarrow \cos ^{-1}\left[\frac{-\left(1-x^{2}\right)}{1+x^{2}}\right]+\frac{1}{2} \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\frac{2 \pi}{3} \end{aligned}                                                    $[\cos (-\theta)=\pi-\cos \theta]$

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

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

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

$\Rightarrow x=\sqrt{3}$                                                                        $\left[\tan \left(\frac{x}{3}\right)=\sqrt{3}\right]$

Hence $x=\sqrt{3}$ is required solution.