Chapter 6 - Analytic Trigonometry - Section 6.2 The Inverse Trigonometric Functions (Continued) - 6.2 Assess Your Understanding - Page 480: 32

$-\frac{\sqrt {2}}{2}$.

1. Let $cos^{-1}(-\frac{\sqrt 3}{3})=t$, we have $cos(t)=-\frac{\sqrt 3}{3}$ and $t$ in quadrant II. 2. Let $x=-\sqrt 3, r=3$, we have $y=\sqrt {3^2-(-\sqrt 3)^2}=\sqrt {6}$. 3. Thus $cot(cos^{-1}(-\frac{\sqrt 3}{3}))=cot(t)=\frac{x}{y}=-\frac{\sqrt {2}}{2}$.