#### Please Solve RD Sharma Class 12 Chapter Inverse Trigonometric Function Exercise 3.6 Question 2 Maths Textbook Solution.

Answer:$R - \left \{ n\pi , n \: \epsilon \: Z \right \}$
Hints: First we will write the domain of $\cot\: x$ and $\cot^{-1}\: x$ respectively. After that we will solve for common domain
Given: $f\! (x) = \cot \: x + \cot ^{-1} (x)$
Solution:
Let us first solve for $cot\: x$
$\because$ $\cot \: x$ is defined for  $x \: \epsilon \: R - \left \{ n\pi, n \: \epsilon \: z \right \}$
$\because$ Domain of  $\cot\: x = R - \left \{ n\pi, n \: \epsilon \: z \right \}$                                                                        $\cdot \cdot \cdot \cdot 1$
Let us solve for $\cot^{-1}\left ( x \right )$
$\because$$\cot^{-1}x$ is defined for $x \: \epsilon \: R$
$\because$ Domain of  $\cot^{-1}(x) = R$                                                                                          $\cdot \cdot \cdot \cdot 2$
Now to find the domain of $f\! (x)$  we take the intersection of both equations 1 and 2
(Domain of $\cot\: x$) $\cap$ (Domain of $\cot ^{-1}x$)
$= R - \left \{ n\pi, n \: \epsilon \: z \right \}$
Hence the domain of $f\! (x) = \cot\: x + \cot^{-1}x$  is  $R - \left \{ n\pi, n \: \epsilon \: z \right \}$