Problem
Evaluate: \( \csc\left(\cot^{-1}\left(\frac{-12}{5}\right)\right) \)
Solution
Let \( \theta = \cot^{-1}\left(\frac{-12}{5}\right) \)
Then:
\[ \cot \theta = \frac{-12}{5} = \frac{\text{Base}}{\text{Perpendicular}} \]
- Base = -12
- Perpendicular = 5
Hypotenuse:
\[ \sqrt{(-12)^2 + 5^2} = \sqrt{144 + 25} = 13 \]
Now,
\[ \sin \theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{5}{13} \]
\[ \csc \theta = \frac{1}{\sin \theta} = \frac{13}{5} \]
Therefore:
\[ \csc\left(\cot^{-1}\left(\frac{-12}{5}\right)\right) = \frac{13}{5} \]
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 sine is positive.