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

$x=\pm \sqrt{\frac{7}{2}}$
Hint:
Here, we will use the formula
$\tan^{-1}A+\tan^{-1}B=\tan^{-1}\left ( \frac{A+B}{1-AB} \right )$
Given:
$\tan^{-1}\left ( \frac{x-2}{x-1} \right )+\tan^{-1}\left ( \frac{x+2}{x+1} \right )=\frac{\pi }{4}$
Solution:
\begin{aligned} &\Rightarrow \tan ^{-1}\left[\frac{\left(\frac{x-2}{x-1}\right)+\left(\frac{x+2}{x+1}\right)}{1-\left(\frac{x-2}{x-1}\right)\left(\frac{x+2}{x+1}\right)}\right]=\frac{\pi}{4}\\ &\Rightarrow \tan ^{-1}\left[\frac{\frac{x-2}{x-1}+\frac{x+2}{x+1}}{1-\left(\frac{x^{2}-2^{2}}{x^{2}-1^{2}}\right)}\right]=\frac{\pi}{4}\\ &\Rightarrow \tan ^{-1}\left[\frac{\frac{(x-2)(x+1)+(x+2)(x-1)}{x^{2}-1}}{\frac{x^{2}-1-x^{2}+4}{x^{2}-1}}\right]=\frac{\pi}{4}\\ &\Rightarrow\left[\frac{\frac{(x-2)(x+1)+(x+2)(x-1)}{x^{2}-1}}{\frac{x^{2}-1-x^{2}+4}{x^{2}-1}}\right]=\tan \frac{\pi}{4}\\ \end{aligned}$\left [ \tan\frac{\pi }{4}= 1 \right ]$
$\! \! \! \! \! \! \! \! \! \Rightarrow\frac{x^{2}+x-2x-2+x^{2}-x+2x-2}{ 3}=1\\ \Rightarrow 2x^{2}-4=3\\ \Rightarrow 2x^{2}=7\\ \Rightarrow x^{2}=\frac{7}{2}\\\Rightarrow x= \pm \sqrt{\frac{7}{2}}$