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

Let ${\sin ^{ - 1}}\frac{5}{{13}} = x\;\;and\;\;{\cos ^{ - 1}}\frac{3}{5} = y$
$\Rightarrow$ $\sin x = \frac{5}{{13}}\;\;and\;\;\cos y = \frac{3}{5}\;\;or,\;\;\tan x = \frac{5}{{12}}\;\;and\;\;\tan y = \frac{4}{3}$

$\Rightarrow$ $x + y = {\tan ^{ - 1}}\frac{5}{{12}} + {\tan ^{ - 1}}\frac{4}{3} = {\tan ^{ - 1}}\left( {\frac{{\frac{5}{{12}} + \frac{4}{3}}}{{1 - \frac{5}{{12}} \times \frac{4}{3}}}} \right) = {\tan ^{ - 1}}\left( {\frac{{\frac{{5 + 16}}{{12}}}}{{\frac{4}{9}}}} \right)$

$= {\tan ^{ - 1}}\left( {\frac{{21}}{{12}} \times \frac{9}{4}} \right) = {\tan ^{ - 1}}\frac{{63}}{{16}}$