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

$x=\pm\sqrt{2}$
Hint:
Here, we will use the formula
$\tan^{-1}A+\tan^{-1}B=\tan^{-1}\left ( \frac{A+B}{1-AB} \right )$
Given:
$\tan^{-1}\left ( \frac{x-2}{x-4} \right )+\tan^{-1}\left ( \frac{x+2}{x+4} \right )=\frac{\pi }{4}$
Solution:
\begin{aligned} &\tan ^{-1}\left[\frac{\left(\frac{x-2}{x-4}\right)+\left(\frac{x+2}{x+4}\right)}{1-\left(\frac{x-2}{x-4}\right)\left(\frac{x+2}{x+4}\right)}\right]=\frac{\pi}{4} \\ &\Rightarrow \tan ^{-1}\left[\frac{\frac{x-2}{x-4}+\frac{x+2}{x+4}}{1-\left(\frac{x^{2}-2^{2}}{x^{2}-4^{2}}\right)}\right]=\frac{\pi}{4} \\ &\Rightarrow \frac{x^{2}+4 x-2 x-8+x^{2}-4 x+2 x-8}{x^{2}-16-x^{2}+4}=\tan \frac{\pi}{4} \\ &\Rightarrow \frac{2 x^{2}-16}{-12}=\tan \frac{\pi}{4} \\ &\Rightarrow \frac{2 x^{2}-16}{-12}=1 \quad\left[\because \tan \frac{\pi}{4}=1\right] \end{aligned}
$\! \! \! \! \! \! \! \! \! \Rightarrow 2x^{2}-16=-12\\ \Rightarrow 2x^{2}=16-12\\ \Rightarrow 2x^{2}=4\\ \Rightarrow x^{2}=2\\ \Rightarrow x=\pm \sqrt2\\$