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

$x=\left ( \frac{-1}{6} \right ) \: or \: x= 1$
For solving this, we can use the formula of union trigonometric function,
$\tan^{-1}A+\tan^{-1}B=\tan^{-1}\left ( \frac{A+B}{1-AB } \right )$
Given:
$\tan^{-1}2x+\tan^{-1}3x=nx+\frac{3\pi }{4}$    and we have to find the value of x.
Solution:
Here A=2x  and  B=3x
So, let’s put the values of A and B in the formula of$\tan^{-1}A+ \tan^{-1}B$
\begin{aligned} &\Rightarrow \tan ^{-1} 2 x+\tan ^{-1} 3 x=n x+\frac{3 \pi}{4} \\ &\Rightarrow \tan ^{-1}\left(\frac{2 x+3 x}{1-(2 x)(3 x)}\right)=n x+\frac{3 \pi}{4} \end{aligned}
\begin{aligned} &\Rightarrow \tan ^{-1}\left(\frac{5 x}{1-6 x^{2}}\right)=n \pi+\frac{3 \pi}{4} \\ &\Rightarrow \frac{5 x}{1-6 x^{2}}=\tan \left(n \pi+\frac{3 \pi}{4}\right) \\ &\Rightarrow \frac{5 x}{1-6 x^{2}}=\frac{\tan (n \pi)+\tan \frac{3 \pi}{4}}{1-\tan (n \pi) \tan \frac{3 \pi}{4}} \end{aligned} \quad\left [\because \operatorname{tann} \pi=\mathbf{0}, \tan \frac{3\pi }{4}= -1 \right ]
\begin{aligned} &\Rightarrow \frac{5 x}{1-6 x^{2}}=-1 \\ &\Rightarrow 5 x=6 x^{2}-1 \\ &\Rightarrow 6 x^{2}-6 x+x-1=0 \\ &\Rightarrow 6 x(x-1)+1(x-1)=0 \\ &\Rightarrow(6 x+1)(x-1)=0 \end{aligned}
$\Rightarrow Either\: 6x+1=0 \: or \: x -1=0$
$\Rightarrow 6x=-1 \: or \: x=1$
$\Rightarrow x=\frac{-1}{6} , 1$