#### Please Solve RD Sharma Class 12 Chapter Inverse Trigonometric Function Exercise 3.11 Question 3 Subquestion (ii) Maths Textbook Solution.

$x=\frac{1}{4},-8$
Hint:
Here, we can use the below formula,
$\tan^{-1}A+\tan^{-1}B=\tan^{-1}\left ( \frac{A+B}{1-AB} \right )$
Given:
$\tan^{-1}\left ( x+1 \right )+\tan^{-1}\left (x-1 \right )=\tan^{-1}\left (\frac{8}{31} \right )$
Solution:
Here,     A=x+1
B=x-1
So, let’s put the values of A and B in the formula of $\tan^{-1}A+\tan^{-1}B$
$\! \! \! \! \! \! \! \! \! \Rightarrow \tan^{-1}\left (x+1 \right )+\tan^{-1}\left ( x-1\right ) = \tan^{-1}\left ( \frac{8}{31} \right )\\ \Rightarrow \tan^{-1}\left ( \frac{x+1+x-1}{1-\left (x+1 \right )\left ( x-1 \right )} \right )=\tan^{-1}\frac{8}{31}\\ \Rightarrow \tan^{-1}\left ( \frac{2x}{1-\left ( x^{2}-1 \right )} \right )=\tan^{-1}\frac{8}{31}\\ \Rightarrow \frac{2x}{1-x^{2}+1}=\frac{8}{31}\\ \Rightarrow 31x=8-4x^{2}$
$\! \! \! \! \! \! \! \! \! \Rightarrow 4x^{2}+31x-8=0\\ \Rightarrow 4x^{2}+32x-x-8=0\\ \Rightarrow 4x\left ( x+8 \right )-1\left ( x+8 \right )=0\\ \Rightarrow 4x-1=0 \: or \: x+8=0\\ \Rightarrow x=\frac{1}{4} \: or \: x=- 8\\ \Rightarrow x=\frac{1}{4},-8$