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

$x=\left ( \frac{-1}{2} \right ),\left ( \frac{1}{2} \right )$
Hint:
Here, 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:
We have to solve$\tan^{-1}\left ( x-1 \right )+ \tan^{-1}x+ \tan^{-1}\left ( x+1 \right )= \tan^{-1}3x$  and find the value of x.
Solution:
L.H.S=R.H.S
$\tan^{-1}\left ( x-1 \right )+ \tan^{-1}x+ \tan^{-1}\left ( x+1 \right )= \tan^{-1}3x$
Let’s use in sum formula,
\begin{aligned} &\tan ^{-1}\left(\frac{x-1+x+1}{1-\left(x^{2}-1\right)}\right)+\tan ^{-1} x=\tan ^{-1} 3 x \\ &\Rightarrow \tan ^{-1}\left(\frac{2 x}{2-x^{2}}\right)=\tan ^{-1} 3 x-\tan ^{-1} x \\ &\Rightarrow \tan ^{-1}\left(\frac{2 x}{2-x^{2}}\right)=\tan ^{-1}\left(\frac{3 x-x}{1+3 x^{2}}\right) \\ &\Rightarrow \frac{2 x}{2-x^{2}}=\frac{3 x-x}{1+3 x^{2}} \end{aligned}
\begin{aligned} &\Rightarrow \frac{2 x}{2-x^{2}}=\frac{2 x}{1+3 x^{2}} \\ &\Rightarrow 1+3 x^{2}=2-x^{2} \\ &\Rightarrow 3 x^{2}+x^{2}=2-1 \\ &\Rightarrow 4 x^{2}=1 \\ &\Rightarrow x^{2}=\frac{1}{4} \\ &\Rightarrow x=0, \pm \frac{1}{2} \end{aligned}
$x=\left ( \frac{-1}{2} \right ),\left ( \frac{1}{2} \right ), 0$

Note: We must know the formula of Intersection.
$\tan^{-1}A+\tan^{-1}B=\tan^{-1}\left ( \frac{A+B}{1-AB} \right )$