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#### Explain Solution R.D.Sharma Class 12 Chapter 3 Inverse Trigonometric Function Exercise 3.14 Question 8 Sub Question 3 maths Textbook Solution.

Answer: $x=\frac{1}{\sqrt{3}}$

Given: $\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)+\cot ^{-1}\left(\frac{1-x^{2}}{2 x}\right)=\frac{2 \pi}{3}, x>0$

Hint: Use

$\text { 1. } \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=2 \tan ^{-1} x \text { and }$

$\text { 2. } \cot ^{-1}\left(\frac{1-x^{2}}{2 x}\right)=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=2 \tan ^{-1} x$

Solution: We have,

$\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)+\cot ^{-1}\left(\frac{1-x^{2}}{2 x}\right)=\frac{2 \pi}{3}$

We know that,

$\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=2 \tan ^{-1} x$ and convert $\cot ^{-1}$into $\tan ^{-1}$

$\Rightarrow 2 \tan ^{-1} x+\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\frac{2 \pi}{3}$                                $\left[\cot \theta=\frac{1}{\tan \theta}\right]$

$\Rightarrow 2 \tan ^{-1} x+2 \tan ^{-1} x=\frac{2 \pi}{3}$                                               $\left[\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=2 \tan ^{-1} x\right]$

\begin{aligned} &\Rightarrow 4 \tan ^{-1} x=\frac{2 \pi}{3} \\ &\Rightarrow \tan ^{-1} x=\frac{2 \pi}{12} \\ &\Rightarrow x=\tan \frac{\pi}{6} \end{aligned}                                                                   $\left[\tan ^{-1} x=\frac{\pi}{6} \Rightarrow x=\tan \frac{\pi}{6}\right]$

$\Rightarrow x=\frac{1}{\sqrt{3}}$                                                                                $\left[\tan \frac{\pi}{6}=\frac{1}{\sqrt{3}}\right]$