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

$x=\sqrt{3}$
Hint:
Here, we use the formula of Intersection
$\tan^{-1}A-\tan^{-1}B=\tan^{-1}\left ( \frac{A-B}{1+AB} \right )$
Given:
$\cot ^{-1}x-\cot ^{-1}\left ( x+2 \right )=\frac{\pi }{12}$
Solution:
We know that,
$cot^{-1}x=\tan^{-1}\left ( \frac{1}{x} \right )$
Therefore, the given equation converted into
\begin{aligned} &\tan ^{-1}\left(\frac{1}{x}\right)-\tan ^{-1}\left(\frac{1}{x+2}\right)=\frac{\pi}{12} \\ &\Rightarrow \tan ^{-1}\left(\frac{\frac{1}{x}-\frac{1}{x+2}}{1+\left(\frac{1}{x}\right)\left(\frac{1}{x+2}\right)}\right)=\frac{\pi}{12} \\ &\Rightarrow\left(\frac{\frac{x+2-x}{x(x+2)}}{\frac{x(x+2)+1}{x(x+2)}}\right)=\tan \frac{\pi}{12} \\ &\Rightarrow \frac{2}{x^{2}+2 x+1}=\tan \left(\frac{\pi}{3}-\frac{\pi}{4}\right) \end{aligned}

$\Rightarrow \frac{2}{x^{2}+2x+1} = \tan \left ( \frac{\pi }{3} -\frac{\pi }{4}\right )$                                                                                 $\left [ \because \frac{\pi }{3}-\frac{\pi }{4}= \frac{\pi }{12} \right ]$
\begin{aligned} &\Rightarrow \frac{2}{x^{2}+2 x+1}=\frac{\tan \left(\frac{\pi}{3}\right)-\tan \left(\frac{\pi}{4}\right)}{1+\left(\tan \left(\frac{\pi}{3}\right)\right)\left(\tan \left(\frac{\pi}{4}\right)\right)} \\ &\Rightarrow \frac{2}{x^{2}+2 x+1}=\frac{\sqrt{3}-1}{1+\sqrt{3}} \end{aligned}
$\left.\Rightarrow \frac{2}{x^{2}+2 x+1}=\frac{\sqrt{3}-1}{1+\sqrt{3}} \times \frac{\sqrt{3}+1}{\sqrt{3}+1} \quad \text { multiply and divide by }(\sqrt{3}+1)\right]$
\begin{aligned} &\Rightarrow \frac{2}{(x+1)^{2}}=\frac{3-1}{(1+\sqrt{3})^{2}} \\ &\Rightarrow \frac{2}{(x+1)^{2}}=\frac{2}{(1+\sqrt{3})^{2}} \\ &\Rightarrow(\sqrt{3}+1)^{2}=(x+1)^{2} \\ &\Rightarrow x+1=\sqrt{3}+1 \\ &\Rightarrow x=\sqrt{3} \end{aligned}