Problem
Prove: \( \sin^{-1}\left(\frac{5}{13}\right) + \cos^{-1}\left(\frac{3}{5}\right) = \tan^{-1}\left(\frac{63}{16}\right) \)
Solution
Let:
\[ A = \sin^{-1}\left(\frac{5}{13}\right), \quad B = \cos^{-1}\left(\frac{3}{5}\right) \]
Step 1: Find tan A
\[ \sin A = \frac{5}{13} \Rightarrow \cos A = \frac{12}{13} \Rightarrow \tan A = \frac{5}{12} \]
Step 2: Find tan B
\[ \cos B = \frac{3}{5} \Rightarrow \sin B = \frac{4}{5} \Rightarrow \tan B = \frac{4}{3} \]
Step 3: Use tan(A + B)
\[ \tan(A+B) = \frac{\tan A + \tan B}{1 – \tan A \tan B} \]
\[ = \frac{\frac{5}{12} + \frac{4}{3}}{1 – \frac{5}{12}\cdot\frac{4}{3}} \]
\[ = \frac{\frac{5}{12} + \frac{16}{12}}{1 – \frac{20}{36}} = \frac{\frac{21}{12}}{\frac{16}{36}} \]
\[ = \frac{21}{12} \cdot \frac{36}{16} = \frac{63}{16} \]
Step 4: Conclude
Since \( A+B \in (-\frac{\pi}{2}, \frac{\pi}{2}) \),
\[ A + B = \tan^{-1}\left(\frac{63}{16}\right) \]
Final Result
\[ \boxed{\tan^{-1}\left(\frac{63}{16}\right)} \]
Explanation
Using triangle values and tan(A+B) identity proves the result.