Problem
Prove: \( \sin^{-1}\left(\frac{63}{65}\right) = \sin^{-1}\left(\frac{5}{13}\right) + \cos^{-1}\left(\frac{3}{5}\right) \)
Solution
Let:
\[ A = \sin^{-1}\left(\frac{5}{13}\right), \quad B = \cos^{-1}\left(\frac{3}{5}\right) \]
Step 1: Find sin A, cos A
\[ \sin A = \frac{5}{13} \Rightarrow \cos A = \frac{12}{13} \]
Step 2: Find sin B, cos B
\[ \cos B = \frac{3}{5} \Rightarrow \sin B = \frac{4}{5} \]
Step 3: Use sin(A + B)
\[ \sin(A+B) = \sin A \cos B + \cos A \sin B \]
\[ = \frac{5}{13}\cdot\frac{3}{5} + \frac{12}{13}\cdot\frac{4}{5} \]
\[ = \frac{15}{65} + \frac{48}{65} = \frac{63}{65} \]
Step 4: Conclude
Since \( A, B \in [0, \frac{\pi}{2}] \), their sum is also in principal range.
\[ A + B = \sin^{-1}\left(\frac{63}{65}\right) \]
Final Result
\[ \boxed{\sin^{-1}\left(\frac{63}{65}\right)} \]
Explanation
Using sin(A+B) identity and triangle values proves the result.