Problem
Prove: \( \tan^{-1}\left(\frac{1}{7}\right) + \tan^{-1}\left(\frac{1}{13}\right) = \tan^{-1}\left(\frac{2}{9}\right) \)
Solution
Let:
\[ A = \tan^{-1}\left(\frac{1}{7}\right), \quad B = \tan^{-1}\left(\frac{1}{13}\right) \]
Step 1: Use tan(A + B) identity
\[ \tan(A + B) = \frac{\tan A + \tan B}{1 – \tan A \tan B} \]
Step 2: Substitute values
\[ = \frac{\frac{1}{7} + \frac{1}{13}}{1 – \frac{1}{7}\cdot\frac{1}{13}} \]
\[ = \frac{\frac{13 + 7}{91}}{1 – \frac{1}{91}} = \frac{\frac{20}{91}}{\frac{90}{91}} \]
\[ = \frac{20}{90} = \frac{2}{9} \]
\[ \tan(A + B) = \frac{2}{9} \]
Step 3: Conclude
Since \( A, B \in (-\frac{\pi}{2}, \frac{\pi}{2}) \), their sum also lies in principal range.
\[ A + B = \tan^{-1}\left(\frac{2}{9}\right) \]
Final Result
\[ \boxed{\tan^{-1}\left(\frac{2}{9}\right)} \]
Explanation
Using tan(A+B) identity and simplifying gives the required result.