Problem
Prove: \( \tan^{-1}\left(\frac{1}{4}\right) + \tan^{-1}\left(\frac{2}{9}\right) = \sin^{-1}\left(\frac{1}{\sqrt{5}}\right) \)
Solution
Let:
\[ A = \tan^{-1}\left(\frac{1}{4}\right), \quad B = \tan^{-1}\left(\frac{2}{9}\right) \]
Step 1: Use tan(A + B)
\[ \tan(A + B) = \frac{\tan A + \tan B}{1 – \tan A \tan B} \]
\[ = \frac{\frac{1}{4} + \frac{2}{9}}{1 – \frac{1}{4}\cdot\frac{2}{9}} \]
\[ = \frac{\frac{9 + 8}{36}}{1 – \frac{2}{36}} = \frac{\frac{17}{36}}{\frac{34}{36}} = \frac{1}{2} \]
\[ \tan(A + B) = \frac{1}{2} \]
Step 2: Find sin(A + B)
Construct triangle:
- Perpendicular = 1
- Base = 2
Hypotenuse:
\[ \sqrt{1^2 + 2^2} = \sqrt{5} \]
\[ \sin(A + B) = \frac{1}{\sqrt{5}} \]
Step 3: Conclude
Since \( A + B \in (-\frac{\pi}{2}, \frac{\pi}{2}) \),
\[ A + B = \sin^{-1}\left(\frac{1}{\sqrt{5}}\right) \]
Final Result
\[ \boxed{\sin^{-1}\left(\frac{1}{\sqrt{5}}\right)} \]
Explanation
Using tan(A+B) identity and converting to sine using triangle ratios.