Prove

Let $\sin ^{-1} \frac{3}{5}=x$. Then, $\sin x=\frac{3}{5} \Rightarrow \cos x=\sqrt{1-\left(\frac{3}{5}\right)^{2}}=\sqrt{\frac{16}{25}}=\frac{4}{5}$.

$\therefore \tan x=\frac{3}{4} \Rightarrow x=\tan ^{-1} \frac{3}{4}$

$\therefore \sin ^{-1} \frac{3}{5}=\tan ^{-1} \frac{3}{4}$.....(1)

Now, let $\cos ^{-1} \frac{12}{13}=y$. Then, $\cos y=\frac{12}{13} \Rightarrow \sin y=\frac{5}{13}$.

$\therefore \tan y=\frac{5}{12} \Rightarrow y=\tan ^{-1} \frac{5}{12}$

$\therefore \cos ^{-1} \frac{12}{13}=\tan ^{-1} \frac{5}{12}$.....(2)

Let $\sin ^{-1} \frac{56}{65}=z$. Then, $\sin z=\frac{56}{65} \Rightarrow \cos z=\frac{33}{65}$.

$\therefore \tan z=\frac{56}{33} \Rightarrow z=\tan ^{-1} \frac{56}{33}$

$\therefore \sin ^{-1} \frac{56}{65}=\tan ^{-1} \frac{56}{33}$....(3)

Now, we have:

L.H.S. $=\cos ^{-1} \frac{12}{13}+\sin ^{-1} \frac{3}{5}$

$=\tan ^{-1} \frac{5}{12}+\tan ^{-1} \frac{3}{4} \quad[$ Using $(1)$ and $(2)]$

$=\tan ^{-1} \frac{\frac{5}{12}+\frac{3}{4}}{1-\frac{5}{12} \cdot \frac{3}{4}} \quad\left[\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1} \frac{x+y}{1-x y}\right]$

$=\tan ^{-1} \frac{20+36}{48-15}$

$=\tan ^{-1} \frac{56}{33}$

$=\sin ^{-1} \frac{56}{65}=$ R.H.S. $\quad[$ Using (3) $]$