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

$x= \left ( \frac{1}{12} \right ),\left ( \frac{-1}{2} \right )$
Hint:
Here, we use the below formula
$\tan^{-1}A+\tan^{-1}B=\tan^{-1}\left ( \frac{A+B}{1-AB} \right )$
Given:
$\tan^{-1}4x+\tan^{-1}6x=\frac{\pi }{4}$
Solution:
Here      A=4x
B=6x
Let’s put the values of A and B in formula of$\tan^{-1}A+\tan^{-1}B$ ,we get
$\tan^{-1}\left ( \frac{4x+6x}{1-(4x)(6x)} \right )=\frac{\pi }{4}$
$\Rightarrow \tan^{-1}\left ( \frac{10x}{1-24x^{2}} \right )=\frac{\pi }{4}$
$\Rightarrow \frac{10x}{1-24x^{2}}=\tan \frac{\pi }{4}$
$\Rightarrow \frac{10x}{1-24x^{2}}=1$                                                                                        $\left ( \tan \left ( \frac{\pi }{4} \right )= 1 \right )$
$\! \! \! \! \! \! \Rightarrow 10x=1-24x^{2}\\ \Rightarrow 24x^{2}+10x-1=0\\ \Rightarrow 24x^{2}+12x-2x-1=0\\ \Rightarrow 12x\left ( 2x+1 \right )-1\left ( 2x+1 \right )=0\\ \Rightarrow \left ( 12x-1 \right )\left ( 2x+1 \right )=0\\$
$\! \! \! \! \! \! \! \! \! \Rightarrow Either\: 12x-1=0 \: or \: 2x+1=0\\ \Rightarrow 12x=1 \: or \: 2x= -1 \\ \Rightarrow x=\frac{1}{12} \: or \: x =-\frac{1}{2}\\ \Rightarrow x=\frac{1}{12},\left ( \frac{-1}{2} \right )\\$