Problem
Evaluate: \( \cot^{-1}(\cot \frac{19\pi}{6}) \)
Solution
First, reduce the angle:
\[ \frac{19\pi}{6} = 3\pi + \frac{\pi}{6} \]
Since cotangent has period \( \pi \):
\[ \cot \frac{19\pi}{6} = \cot \frac{\pi}{6} \]
Now,
\[ \cot \frac{\pi}{6} = \sqrt{3} \]
Thus the expression becomes:
\[ \cot^{-1}(\sqrt{3}) \]
Recall the principal value range of \( \cot^{-1} x \):
\[ (0, \pi) \]
We need an angle in this range whose cotangent is \( \sqrt{3} \).
We know that:
\[ \cot \frac{\pi}{6} = \sqrt{3} \]
And \( \frac{\pi}{6} \) lies in the principal value range.
Final Answer
\[ \boxed{\frac{\pi}{6}} \]