Prove $\tan ^{-1} \frac{63}{16}=\sin ^{-1} \frac{5}{13}+\cos ^{-1} \frac{3}{5}$

Let $\sin ^{-1} \frac{5}{13} = x$.
Then,$\sin x = \frac{5}{13}$,which implies $\cos x = \sqrt{1 - (\frac{5}{13})^2} = \sqrt{1 - \frac{25}{169}} = \sqrt{\frac{144}{169}} = \frac{12}{13}$.
Therefore,$\tan x = \frac{\sin x}{\cos x} = \frac{5/13}{12/13} = \frac{5}{12}$,so $x = \tan ^{-1} \frac{5}{12}$.
Thus,$\sin ^{-1} \frac{5}{13} = \tan ^{-1} \frac{5}{12}$ $\ldots (1)$.
Let $\cos ^{-1} \frac{3}{5} = y$.
Then,$\cos y = \frac{3}{5}$,which implies $\sin y = \sqrt{1 - (\frac{3}{5})^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$.
Therefore,$\tan y = \frac{\sin y}{\cos y} = \frac{4/5}{3/5} = \frac{4}{3}$,so $y = \tan ^{-1} \frac{4}{3}$.
Thus,$\cos ^{-1} \frac{3}{5} = \tan ^{-1} \frac{4}{3}$ $\ldots (2)$.
Now,consider the $R$.$H$.$S$.:
$\sin ^{-1} \frac{5}{13} + \cos ^{-1} \frac{3}{5} = \tan ^{-1} \frac{5}{12} + \tan ^{-1} \frac{4}{3}$.
Using the formula $\tan ^{-1} a + \tan ^{-1} b = \tan ^{-1} (\frac{a+b}{1-ab})$:
$= \tan ^{-1} (\frac{\frac{5}{12} + \frac{4}{3}}{1 - \frac{5}{12} \times \frac{4}{3}}) = \tan ^{-1} (\frac{\frac{15+48}{36}}{1 - \frac{20}{36}}) = \tan ^{-1} (\frac{63/36}{16/36}) = \tan ^{-1} \frac{63}{16}$.
$= L.H.S.$
Hence,the identity is proved.