#### Explain Solution R.D.Sharma Class 12 Chapter 3 Inverse Trigonometric Function Exercise 3.14 Question 8 Sub Question 6 maths Textbook Solution.

Answer: $x=\sqrt{3}$

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

Hint: Using formula

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

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

Solution:

We have,

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

We know that,

\begin{aligned} &\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)=2 \tan ^{-1} x \\ &\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=2 \tan ^{-1} x \\ &\Rightarrow \cos ^{-1}\left[\frac{-\left(1-x^{2}\right)}{1+x^{2}}\right]+\frac{1}{2} \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\frac{2 \pi}{3} \end{aligned}                                                $[\cos (-\theta)=\pi-\cos \theta]$

\begin{aligned} &\Rightarrow \pi-\cos ^{-1}\left[\frac{\left(1-x^{2}\right)}{1+x^{2}}\right]+\frac{1}{2} \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\frac{2 \pi}{3} \\ &\Rightarrow \pi-2 \tan ^{-1}(x)+\frac{1}{2} \times 2 \tan ^{-1} x=\frac{2 \pi}{3} \end{aligned}

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

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

Hence $x=\sqrt{3}$ is required solution.

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Answer: $x=\pm \sqrt{\frac{7}{2}}$

Given: $\tan ^{-1}\left(\frac{x-2}{x-1}\right)+\tan ^{-1}\left(\frac{x+2}{x+1}\right)=\frac{\pi}{4}$

Hint: Use $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$

Solution: We have,

$\tan ^{-1}\left(\frac{x-2}{x-1}\right)+\tan ^{-1}\left(\frac{x+2}{x+1}\right)=\frac{\pi}{4}$

We know that,

$\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$

\begin{aligned} &\Rightarrow \tan ^{-1}\left[\frac{\frac{x-2}{x-1}+\frac{x+2}{x+1}}{1-\frac{x-2}{x-1} \times \frac{x+2}{x+1}}\right]=\frac{\pi}{4} \\ &\Rightarrow \tan ^{-1}\left[\frac{\frac{(x-2)(x+1)+(x-1)(x+2)}{\frac{x^{2}-1-[(x-2)(x+2)]}{x^{2}-1}}}{\frac{x^{2}-1}{4}}=\frac{\pi}{4}\right. \\ &\Rightarrow \tan ^{-1}\left[\frac{x^{2}+x-2 x-2+x^{2}+2 x-x-2}{x^{2}-1-x^{2}+4}\right]=\frac{\pi}{4} \end{aligned}

$\Rightarrow \tan ^{-1}\left[\frac{2 x^{2}-4}{3}\right]=\tan ^{-1}(1)$                                                                            $\left[\tan ^{-1}(1)=\frac{\pi}{4}\right]$

\begin{aligned} &\Rightarrow \frac{2 x^{2}-4}{3}=1 \\ &\Rightarrow 2 x^{2}=3+4 \\ &\Rightarrow 2 x^{2}=7 \end{aligned}

\begin{aligned} &\Rightarrow x^{2}=\frac{7}{2} \\ &\Rightarrow x=\pm \sqrt{\frac{7}{2}} \end{aligned}

Hence,  $x=\pm \sqrt{\frac{7}{2}}$ is required solution