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

Answer: $\frac{\pi }{6}$
Hint: The $cot^{-1}$ function is defined as a function whose domain 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 $y = cot^{-1}(\sqrt{3})$                                                                                                     $\cdot \cdot \cdot 1$
$cot\: y = \sqrt{3}$
$cot\: y = cot\frac{\pi }{6} \; \; \; \; \; \; \; \; \; [cot\frac{\pi }{6}= \sqrt{3} ]$
$y = \frac{\pi }{6}$
$cot^{-1}\sqrt{3}= \frac{\pi }{6}$                                                                         (From equation 1)
$\because$ The principal value branch of the function$cot ^{-1}$ is$(0, \pi )$
$cot^{-1}(\sqrt{3})$$= \frac{\pi }{6} \: \epsilon \: (0, \pi )$
Hence the principal value of $cot^{-1} (\sqrt{3})$ is  $\frac{\pi }{6}$