Problem
Evaluate: \( \sin\left(\cos^{-1}\left(\frac{-3}{5}\right) + \cot^{-1}\left(\frac{-5}{12}\right)\right) \)
Solution
Let:
\[ A = \cos^{-1}\left(\frac{-3}{5}\right), \quad B = \cot^{-1}\left(\frac{-5}{12}\right) \]
Step 1: Find sin A and cos A
\[ \cos A = \frac{-3}{5} = \frac{\text{Base}}{\text{Hypotenuse}} \]
- Base = -3
- Hypotenuse = 5
Perpendicular:
\[ \sqrt{5^2 – 3^2} = \sqrt{25 – 9} = 4 \]
\[ \sin A = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{4}{5} \]
Step 2: Find sin B and cos B
\[ \cot B = \frac{-5}{12} = \frac{\text{Base}}{\text{Perpendicular}} \]
- Base = -5
- Perpendicular = 12
Hypotenuse:
\[ \sqrt{(-5)^2 + 12^2} = 13 \]
\[ \sin B = \frac{12}{13}, \quad \cos B = \frac{-5}{13} \]
Step 3: Use sin(A + B) formula
\[ \sin(A + B) = \sin A \cos B + \cos A \sin B \]
\[ = \frac{4}{5} \cdot \frac{-5}{13} + \frac{-3}{5} \cdot \frac{12}{13} \]
\[ = \frac{-20}{65} + \frac{-36}{65} = \frac{-56}{65} \]
Final Answer
\[ \boxed{-\frac{56}{65}} \]
Explanation
Angles lie in second quadrant (for cos⁻¹ negative) and second quadrant (for cot⁻¹ negative), so signs are handled accordingly.