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

$x=\left (-20 \right ), \frac{1}{4}$
Hint:
Here, we use the formula
$\tan^{-1}A+\tan^{-1}B=\tan^{-1}\left ( \frac{A+B}{1-AB} \right )$
Given:
$\tan^{-1}\left ( x+2 \right )+\tan^{-1}\left (x-2 \right )=\tan^{-1}\left ( \frac{8}{79} \right )$
Solution:
Here we have$A =x+2\: and\: B=x-2$
Using the value of A and B in the formula of $\tan^{-1}A+\tan^{-1}B$
\begin{aligned} &\tan ^{-1}(x+2)+\tan ^{-1}(x-2)=\tan ^{-1}\left(\frac{8}{79}\right) \\ &\Rightarrow \tan ^{-1}\left[\frac{x+2+x-2}{1-(x+2)(x-2)}\right]=\tan ^{-1}\left(\frac{8}{79}\right) \\ &\Rightarrow\left[\frac{2 x}{1-x^{2}+4}\right]=\left(\frac{8}{79}\right) \end{aligned}
\begin{aligned} &\Leftrightarrow\left[\frac{2 x}{5-x^{2}}\right]=\left(\frac{8}{79}\right) \\ &\Rightarrow 2 x(79)=8\left(5-x^{2}\right) \\ &\Rightarrow 158 x=40-8 x^{2} \\ &\Rightarrow 8 x^{2}+158 x-40=0 \\ &\Rightarrow 79 x-20=0 \\ &\Rightarrow 4 x^{2}+80 x-x-20=0 \\ &\Rightarrow 4 x(x+20)-1(x+20)=0 \\ &\Rightarrow x+20=0 \text { or } 4 x-1=0 \\ &\Rightarrow x=-20 \text { or } 4 x=1 \end{aligned}
$\Rightarrow x=\left (-20 \right ), 14$