Explanation:
Let's break it down step by step. We have \( \cot^{-1}(\sqrt{3}) \), which means the cotangent of an angle whose tangent is \( \sqrt{3} \).
Now, recall that \( \cot(\theta) = \frac{1}{\tan(\theta)} \) and \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \).
Since \( \tan(\theta) = \sqrt{3} \), we can express \( \cot^{-1}(\sqrt{3}) \) as \( \cot^{-1}\left(\frac{1}{\tan(\theta)}\right) \).
Now, \( \cos(\theta) = \frac{1}{\sec(\theta)} \), and \( \sec(\theta) = \frac{1}{\cos(\theta)} \).
So, \( \cos(\cot^{-1}(\sqrt{3})) = \cos(\cot^{-1}(\frac{1}{\tan(\theta)})) = \cos(\cot^{-1}(\frac{1}{\sqrt{3}})) \).
Now, \( \cot^{-1}(\frac{1}{\sqrt{3}}) \) represents an angle whose cotangent is \( \frac{1}{\sqrt{3}} \). This is equivalent to \( \frac{\pi}{6} \), because \( \cot(\frac{\pi}{6}) = \frac{1}{\tan(\frac{\pi}{6})} = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}} \).
So, \( \cos(\cot^{-1}(\sqrt{3})) = \cos(\frac{\pi}{6}) \).
Now, \( \cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2} \).
Therefore, \( \cos(\cot^{-1}(\sqrt{3})) = \frac{\sqrt{3}}{2} \).