1. **State the problem:** Find the value of $\cot^{-1}(-\sqrt{3})$ in radians. 2. **Recall the definition:** The inverse cotangent function $\cot^{-1}(x)$ gives an angle $\theta$ such that $\cot(\theta) = x$ and $\theta \in (0, \pi)$. 3. **Use the cotangent values:** We know $\cot(\frac{\pi}{6}) = \sqrt{3}$. 4. **Find the angle for negative value:** Since $\cot(\theta) = -\sqrt{3}$, and cotangent is negative in the second quadrant $(\frac{\pi}{2}, \pi)$, the angle is $\pi - \frac{\pi}{6} = \frac{5\pi}{6}$. 5. **Final answer:** $$\cot^{-1}(-\sqrt{3}) = \frac{5\pi}{6}$$