Problem
Prove: \( \sin^{-1}\left(\frac{12}{13}\right) + \cos^{-1}\left(\frac{4}{5}\right) + \tan^{-1}\left(\frac{63}{16}\right) = \pi \)
Solution
Let:
\[ A = \sin^{-1}\left(\frac{12}{13}\right), \quad B = \cos^{-1}\left(\frac{4}{5}\right) \]
Step 1: Find tan A
\[ \sin A = \frac{12}{13} = \frac{\text{Perpendicular}}{\text{Hypotenuse}} \]
- Perpendicular = 12
- Hypotenuse = 13
Base:
\[ \sqrt{13^2 – 12^2} = 5 \]
\[ \tan A = \frac{12}{5} \]
Step 2: Find tan B
\[ \cos B = \frac{4}{5} = \frac{\text{Base}}{\text{Hypotenuse}} \]
- Base = 4
- Hypotenuse = 5
Perpendicular:
\[ \sqrt{5^2 – 4^2} = 3 \]
\[ \tan B = \frac{3}{4} \]
Step 3: Find tan(A + B)
\[ \tan(A + B) = \frac{\tan A + \tan B}{1 – \tan A \tan B} \]
\[ = \frac{\frac{12}{5} + \frac{3}{4}}{1 – \frac{12}{5} \cdot \frac{3}{4}} \]
\[ = \frac{\frac{48 + 15}{20}}{1 – \frac{36}{20}} = \frac{\frac{63}{20}}{\frac{-16}{20}} = -\frac{63}{16} \]
\[ \tan(A + B) = -\frac{63}{16} \]
Step 4: Interpret angle
\[ A + B = \pi – \tan^{-1}\left(\frac{63}{16}\right) \]
Step 5: Final proof
\[ A + B + \tan^{-1}\left(\frac{63}{16}\right) = \pi \]
Final Result
\[ \boxed{\pi} \]
Explanation
Using triangle ratios and tan(A+B) identity, we show the sum equals π.