#### Please Solve RD Sharma Class 12 Chapter Inverse Trigonometric Function Exercise 3.6 Question 1 Subquestion (i) Maths Textbook Solution.

Answer: $\frac{5\pi }{6}$
Hints: The $\cot ^{-1}$ function is defined as a function whose domain is R and the principal value branch of the function $cot^{-1}$ is $\left ( 0,\pi \right )$
Thus,$cot^{-1}: R \rightarrow (0,\pi )$
Given:$cot^{-1}(\sqrt{-3})$
Solution:
Let$y= cot^{-1}(\sqrt{-3}) \; \; \; \; \; \; \cdot \cdot \cdot \left ( 1 \right )$
$cot\: y= -\sqrt{3}$
$cot\: y = - cot\frac{\pi }{6} \; \; \; \; \; \; [cot\:\frac{\pi }{6} =\sqrt{3} ]$
$cot\: y = cot (\pi - \frac{\pi }{6} ) \; \; \; \; \; \; \; \; \; [ \because cot\left ( \pi - \theta \right ) = - cot\theta ]$
$cot\: y = cot\frac{5\pi }{6}$
$y = \frac{5\pi }{6}$$cot^{-1}(\sqrt{-3}) = \frac{5\pi }{6}$               (From equation 1)
$\because$ The principal value branch of the function $cot^{-1} is (0, \pi )$
$cot^{-1}(\sqrt{-3}) = \frac{5\pi }{6} \: \epsilon\: (0, \pi )$
Hence the principal value of $cot^{-1} (\sqrt{-3})$ is  $\frac{5\pi }{6}$