Problem
Evaluate: \( \cot\left(\sec^{-1}\left(\frac{-13}{5}\right)\right) \)
Solution
Let \( \theta = \sec^{-1}\left(\frac{-13}{5}\right) \)
Then:
\[ \sec \theta = \frac{-13}{5} \]
\[ \cos \theta = \frac{1}{\sec \theta} = -\frac{5}{13} \]
Using triangle:
- Base = -5
- Hypotenuse = 13
Perpendicular:
\[ \sqrt{13^2 – 5^2} = \sqrt{169 – 25} = 12 \]
Now,
\[ \cot \theta = \frac{\text{Base}}{\text{Perpendicular}} = \frac{-5}{12} \]
Therefore:
\[ \cot\left(\sec^{-1}\left(\frac{-13}{5}\right)\right) = -\frac{5}{12} \]
Final Answer
\[ \boxed{-\frac{5}{12}} \]
Explanation
Since sec⁻¹x gives angles in the second quadrant for negative values, cosine and cotangent are negative.