Question

\[\cot^{-1} \left( - \frac{1}{\sqrt{3}} \right)\]

Hint:

When finding the principal value of inverse trigonometric functions, ensure that the value lies within the specified range (for \( \cot^{-1} \), the range is \( (0, \pi) \)).

Answer 1

Step 1: Recall that the range of \( \cot^{-1} \) is \( (0, \pi) \), and we need to find the angle whose cotangent is \( -\frac{1}{\sqrt{3}} \). 

Step 2: Since \( \cot \theta = \frac{1}{\tan \theta} \), we have: \[ \cot \theta = -\frac{1}{\sqrt{3}} \quad \Rightarrow \quad \tan \theta = -\sqrt{3}. \] 

Step 3: The principal value of \( \tan^{-1}(-\sqrt{3}) \) is \( -\frac{\pi}{3} \), but since the range of \( \cot^{-1} \) is \( (0, \pi) \), we adjust the angle to: \[ \cot^{-1} \left( -\frac{1}{\sqrt{3}} \right) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}. \] Thus, the principal value is \( \frac{2\pi}{3} \).