Prove that: \(\sin^{-1}\left(\frac{3}{5}\right) + \cos^{-1}\left(\frac{12}{13}\right) = \sin^{-1}\left(\frac{56}{65}\right)\)

Answer: Proved (both angles are acute and \(\sin(\alpha + \beta) = \frac{56}{65}\)).

Let \(\sin^{-1}\frac{3}{5} = \alpha\), so \(\sin\alpha = \frac{3}{5}\), \(\cos\alpha = \frac{4}{5}\), \(0 \lt \alpha \lt \frac{\pi}{2}\).

Let \(\cos^{-1}\frac{12}{13} = \beta\), so \(\cos\beta = \frac{12}{13}\), \(\sin\beta = \frac{5}{13}\), \(0 \lt \beta \lt \frac{\pi}{2}\).

\(\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta = \frac{3}{5}\cdot\frac{12}{13} + \frac{4}{5}\cdot\frac{5}{13} = \frac{36 + 20}{65} = \frac{56}{65}\).

Also \(\cos(\alpha + \beta) = \frac{4}{5}\cdot\frac{12}{13} - \frac{3}{5}\cdot\frac{5}{13} = \frac{33}{65} \gt 0\), so \(0 \lt \alpha + \beta \lt \frac{\pi}{2}\). Hence \(\alpha + \beta = \sin^{-1}\frac{56}{65}\), i.e. \(\sin^{-1}\left(\frac{3}{5}\right) + \cos^{-1}\left(\frac{12}{13}\right) = \sin^{-1}\left(\frac{56}{65}\right)\).