Problem
Prove: \( \cos\left(\sin^{-1}\left(\frac{3}{5}\right) + \cot^{-1}\left(\frac{3}{2}\right)\right) = \frac{6}{5\sqrt{13}} \)
Solution
Let:
\[ A = \sin^{-1}\left(\frac{3}{5}\right), \quad B = \cot^{-1}\left(\frac{3}{2}\right) \]
Step 1: Find sin A and cos A
\[ \sin A = \frac{3}{5} = \frac{\text{Perpendicular}}{\text{Hypotenuse}} \]
- Perpendicular = 3
- Hypotenuse = 5
Base:
\[ \sqrt{5^2 – 3^2} = \sqrt{25 – 9} = 4 \]
\[ \cos A = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{4}{5} \]
Step 2: Find sin B and cos B
\[ \cot B = \frac{3}{2} = \frac{\text{Base}}{\text{Perpendicular}} \]
- Base = 3
- Perpendicular = 2
Hypotenuse:
\[ \sqrt{3^2 + 2^2} = \sqrt{13} \]
\[ \sin B = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{2}{\sqrt{13}}, \quad \cos B = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{3}{\sqrt{13}} \]
Step 3: Use cos(A + B) formula
\[ \cos(A + B) = \cos A \cos B – \sin A \sin B \]
\[ = \frac{4}{5} \cdot \frac{3}{\sqrt{13}} – \frac{3}{5} \cdot \frac{2}{\sqrt{13}} \]
\[ = \frac{12}{5\sqrt{13}} – \frac{6}{5\sqrt{13}} = \frac{6}{5\sqrt{13}} \]
Final Result
\[ \boxed{\frac{6}{5\sqrt{13}}} \]
Explanation
We converted inverse trigonometric expressions into triangle ratios and applied the cos(A+B) identity.