Chapter 6 - Analytic Trigonometry - Section 6.2 The Inverse Trigonometric Functions (Continued) - 6.2 Assess Your Understanding - Page 480: 14

$-\sqrt 3$

The range of $\sin^{-1}{x}$ is $[-\dfrac{\pi}{2},\dfrac{\pi}{2}]$. Here, $\sin^{-1}{(\dfrac{-1}{2})}=\dfrac{-\pi}{6}$, because $\sin{(\dfrac{-\pi}{6})}=\dfrac{-1}{2}$ and $\dfrac{-\pi}{6}$ lies in the range of $\sin^{-1}{x}$ . Therefore, we have: $\cot [\sin^{-1}{(\dfrac{-1}{2})}]=\cot ({\dfrac{-\pi}{6}})=-\sqrt 3$