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

$X = \frac{1}{\sqrt{3}}$
Hint:
Here, we use the formula of Intersection
$\tan^{-1}A-\tan^{-1}B=\tan^{-1}\left ( \frac{A-B}{1+AB} \right )$
Given:
$\tan^{-1}\left ( \frac{1-x}{1+x} \right )-\frac{1}{2}\tan^{-1}x=0$
Solution:
Here let’s assume
\begin{aligned} &x=\tan \theta \\ &\Rightarrow \tan ^{-1}\left(\frac{1-\tan \theta}{1+\tan \theta}\right)-\frac{1}{2} \tan ^{-1}(\tan \theta)=0 \\ &\Rightarrow \tan ^{-1}\left(\frac{\tan \frac{\pi}{4}-\tan \theta}{1+\left(\tan \frac{\pi}{4}\right)(\tan \theta)}\right)-\frac{1}{2} \theta=0 \\ &\Rightarrow \tan ^{-1}\left[\tan \left(\frac{\pi}{4}-\theta\right)\right]-\frac{\theta}{2}=0 \end{aligned}
\begin{aligned} &\Rightarrow \frac{\pi}{4}-\theta-\frac{\theta}{2}=0 \\ &\Rightarrow \frac{3 \theta}{2}=\frac{\pi}{4} \\ &\Rightarrow \theta=\frac{\pi}{6} \\ &\Rightarrow \tan ^{-1} x=\frac{\pi}{6} \\ &\Rightarrow x=\tan \frac{\pi}{6} \\ &x=\frac{1}{\sqrt{3}} \end{aligned}