Problem
Prove: \( \tan\left(\sin^{-1}\left(\frac{5}{13}\right) + \cos^{-1}\left(\frac{3}{5}\right)\right) = \frac{63}{16} \)
Solution
Let:
\[ A = \sin^{-1}\left(\frac{5}{13}\right), \quad B = \cos^{-1}\left(\frac{3}{5}\right) \]
Step 1: Find tan A
\[ \sin A = \frac{5}{13} = \frac{\text{Perpendicular}}{\text{Hypotenuse}} \]
- Perpendicular = 5
- Hypotenuse = 13
Base:
\[ \sqrt{13^2 – 5^2} = \sqrt{169 – 25} = 12 \]
\[ \tan A = \frac{\text{Perpendicular}}{\text{Base}} = \frac{5}{12} \]
Step 2: Find tan B
\[ \cos B = \frac{3}{5} = \frac{\text{Base}}{\text{Hypotenuse}} \]
- Base = 3
- Hypotenuse = 5
Perpendicular:
\[ \sqrt{5^2 – 3^2} = \sqrt{25 – 9} = 4 \]
\[ \tan B = \frac{\text{Perpendicular}}{\text{Base}} = \frac{4}{3} \]
Step 3: Use tan(A + B) formula
\[ \tan(A + B) = \frac{\tan A + \tan B}{1 – \tan A \tan B} \]
\[ = \frac{\frac{5}{12} + \frac{4}{3}}{1 – \frac{5}{12} \cdot \frac{4}{3}} \]
\[ = \frac{\frac{5 + 16}{12}}{1 – \frac{20}{36}} = \frac{\frac{21}{12}}{\frac{16}{36}} \]
\[ = \frac{21}{12} \times \frac{36}{16} = \frac{63}{16} \]
Final Result
\[ \boxed{\frac{63}{16}} \]
Explanation
We converted inverse trigonometric functions into triangle ratios and applied the tan(A+B) identity.