Step-by-step Solution:

\begin{aligned} \text{Step 1: Identify the value of } \sqrt{3} \text{ in terms of cotangent.} \\ \text{We know that:} \\ \cot\left(\frac{\pi}{6}\right) &= \sqrt{3} \\ \text{Thus, we can express } -\sqrt{3} \text{ as:} \\ \cot\left(\frac{\pi}{6}\right) = \sqrt{3} &\implies \cot\left(-\frac{\pi}{6}\right) = -\sqrt{3} \\ \text{Step 2: Use the cotangent identity.} \\ \text{Using the property of cotangent, we have:} \\ \cot(\pi - x) &= -\cot(x) \\ \text{So, we can write:} \\ \cot^{-1}(-\sqrt{3}) &= \cot^{-1}\left(\cot\left(\pi - \frac{\pi}{6}\right)\right) \\ \text{Step 3: Simplify the expression.} \\ \text{Now, we can simplify:} \\ \pi - \frac{\pi}{6} &= \frac{6\pi}{6} - \frac{\pi}{6} = \frac{5\pi}{6} \\ \text{Thus, we have:} \\ \cot^{-1}(-\sqrt{3}) &= \cot^{-1}\left(\cot\left(\frac{5\pi}{6}\right)\right) \\ \text{Step 4: Apply the inverse cotangent function.} \\ \text{Since } \cot^{-1}(\cot(x)) &= x \text{ for } x \text{ in the range of } (0, \pi), \text{ we can conclude:} \\ \cot^{-1}(-\sqrt{3}) &= \frac{5\pi}{6} \end{aligned}