Prove

Let $\sin ^{-1} \frac{5}{13}=x$. Then, $\sin x=\frac{5}{13} \Rightarrow \cos x=\frac{12}{13}$.

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

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

Let $\cos ^{-1} \frac{3}{5}=y$. Then, $\cos y=\frac{3}{5} \Rightarrow \sin y=\frac{4}{5}$.

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

$\therefore \cos ^{-1} \frac{3}{5}=\tan ^{-1} \frac{4}{3}$....(2)

Using (1) and (2), we have

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

$=\tan ^{-1} \frac{5}{12}+\tan ^{-1} \frac{4}{3}$

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

$=\tan ^{-1}\left(\frac{15+48}{36-20}\right)$

$=\tan ^{-1} \frac{63}{16}$

$=$ L.H.S.