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

$x=1,\left ( -6 \right )$
Hint:
Here, we use the below formula
$\tan^{-1}A+\tan^{-1}B=\tan^{-1}\left ( \frac{A+B}{1-AB} \right )$
Given:
$\tan^{-1}\left ( \frac{x}{2} \right )+\tan^{-1}\left ( \frac{x}{3} \right )=\frac{\pi }{4}$ for  $0 < x < \sqrt{6}$
Solution:
$\! \! \! \! \! \! \! \! \! Here,A=\frac{x}{2}\\ B=\frac{x}{2}$
Using the values of A and B in the formula of $\tan^{-1}A+\tan^{-1}B$ , we get
\begin{aligned} &\Rightarrow \tan ^{-1}\left(\frac{x}{2}\right)+\tan ^{-1}\left(\frac{x}{3}\right)=\frac{\pi}{4} \\ &\Rightarrow \tan ^{-1}\left(\frac{\frac{x}{2}+\frac{x}{3}}{1-\left(\frac{x}{2}\right)\left(\frac{x}{3}\right)}\right)=\frac{\pi}{4} \\ &\Rightarrow \tan ^{-1}\left(\frac{\frac{3 x+2 x}{6}}{\frac{6-x^{2}}{6}}\right)=\frac{\pi}{4} \end{aligned}
\begin{aligned} &\Leftrightarrow \frac{5 x}{6-x^{2}}=\tan \frac{\pi}{4} \quad\left[\because \tan \frac{\pi}{4}=1\right] \\ &\Rightarrow 5 x=6-x^{2} \\ &\Rightarrow x^{2}+5 x-6=0 \\ &\Rightarrow x^{2}+6 x-x-6=0 \\ &\Rightarrow x(x+6)-1(x+6)=0 \\ &\Rightarrow(x-1)(x+6)=0 \end{aligned}
$\! \! \! \! \! \! \! \! \! \Rightarrow Either x-1=0 \: or\: x+6=0 \\ \Rightarrow i\! f x-1=0\: then\: x=1 \\ i\! f x+6=0 \: then\: x=-6\\ \Rightarrow x=1,-6$