Problem
Prove: \( \tan\left(\cos^{-1}\left(\frac{4}{5}\right) + \tan^{-1}\left(\frac{2}{3}\right)\right) = \frac{17}{6} \)
Solution
Let:
\[ A = \cos^{-1}\left(\frac{4}{5}\right), \quad B = \tan^{-1}\left(\frac{2}{3}\right) \]
Step 1: Find tan A
\[ \cos A = \frac{4}{5} = \frac{\text{Base}}{\text{Hypotenuse}} \]
- Base = 4
- Hypotenuse = 5
Perpendicular:
\[ \sqrt{5^2 – 4^2} = \sqrt{25 – 16} = 3 \]
\[ \tan A = \frac{\text{Perpendicular}}{\text{Base}} = \frac{3}{4} \]
Step 2: Find tan B
\[ \tan B = \frac{2}{3} = \frac{\text{Perpendicular}}{\text{Base}} \]
- Perpendicular = 2
- Base = 3
Step 3: Use tan(A + B) formula
\[ \tan(A + B) = \frac{\tan A + \tan B}{1 – \tan A \tan B} \]
\[ = \frac{\frac{3}{4} + \frac{2}{3}}{1 – \frac{3}{4} \cdot \frac{2}{3}} \]
\[ = \frac{\frac{9 + 8}{12}}{1 – \frac{6}{12}} = \frac{\frac{17}{12}}{\frac{1}{2}} \]
\[ = \frac{17}{12} \times 2 = \frac{17}{6} \]
Final Result
\[ \boxed{\frac{17}{6}} \]
Explanation
We converted inverse trigonometric functions into triangle ratios and applied the tan(A+B) identity.