Chapter 6 - Analytic Trigonometry - Section 6.2 The Inverse Trigonometric Functions (Continued) - 6.2 Assess Your Understanding - Page 480: 37

$\dfrac{ \pi }{6}$

We know that the trigonometric function $\cot^{-1} x$ has range: $(0, \pi)$. Here, $\cot^{-1} (\sqrt 3)=\dfrac{\pi}{6}$, because $\cot(\dfrac{\pi}{6})=\dfrac{\sqrt 3}{1}$ and $\dfrac{\pi}{6}$ lies in the range of $\cot^{-1}{x}$ .