Let $x=\cos ^{-1} \frac{12}{13} \Rightarrow \cos x=\frac{12}{13}$
$\therefore \sin x=\sqrt{1-\cos ^2 x}=\sqrt{1-\frac{144}{169}}=\sqrt{\frac{25}{169}}=\frac{5}{13} \text { and let } y=\sin ^{-1} \frac{3}{5} \Rightarrow \sin y=\frac{3}{5} \\$
$\therefore \cos y=\sqrt{1-\sin ^2 y}=\sqrt{1-\frac{9}{25}}=\sqrt{\frac{16}{25}}=\frac{4}{5} \\$
$\sin (x+y)=\sin x \cos y+\cos x \sin y=\frac{5}{13} \cdot \frac{4}{5}+\frac{12}{13} \cdot \frac{3}{5}=\frac{20}{65}+\frac{36}{65}=\frac{56}{65} \\$
$\therefore x+y=\sin ^{-1} \frac{56}{65} \text {. Hence, } \cos ^{-1} \frac{12}{13}+\sin ^{-1} \frac{3}{5}=\sin ^{-1} \frac{56}{65}$
