Problem
Evaluate: \( \sec\left(\cot^{-1}\left(\frac{-5}{12}\right)\right) \)
Solution
Let \( \theta = \cot^{-1}\left(\frac{-5}{12}\right) \)
Then:
\[ \cot \theta = \frac{-5}{12} = \frac{\text{Base}}{\text{Perpendicular}} \]
- Base = -5
- Perpendicular = 12
Hypotenuse:
\[ \sqrt{(-5)^2 + 12^2} = \sqrt{25 + 144} = 13 \]
Now,
\[ \cos \theta = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{-5}{13} \]
\[ \sec \theta = \frac{1}{\cos \theta} = \frac{-13}{5} \]
Therefore:
\[ \sec\left(\cot^{-1}\left(\frac{-5}{12}\right)\right) = -\frac{13}{5} \]
Final Answer
\[ \boxed{-\frac{13}{5}} \]
Explanation
Since cot⁻¹x lies in (0, π), a negative value places the angle in the second quadrant where cosine is negative.