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

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

…….(i)
Let ${\sin ^{ - 1}}\frac{5}{{13}} = x$
$\Rightarrow$ $\sin x = \frac{5}{{13}}$
and ${\cos ^2}x = 1 - {\sin ^2}x$

$= 1 - \frac{{25}}{{169}} = \frac{{144}}{{169}}$
$\Rightarrow$ $\cos x = \sqrt {\frac{{144}}{{169}}} = \frac{{12}}{{13}}$

$therefore,$ $\tan x = \frac{{\sin x}}{{\cos x}} = \frac{{5/13}}{{12/13}} = \frac{5}{{12}}$

……(ii)
$\Rightarrow$ $\tan x = 5/12$

…….(iii)
Again, let ${\cos ^{ - 1}}\frac{3}{5} = y \Rightarrow \cos y = \frac{3}{5}$

$therefore,$ $\sin y = \sqrt {1 - {{\cos }^2}y}$
$= \sqrt {1 - {{\left( {\frac{3}{5}} \right)}^2}} = \sqrt {1 - \frac{9}{{25}}}$

$\sin y = \sqrt {\frac{{16}}{{25}}} = \frac{4}{5}$
$\Rightarrow$ $\tan y = \frac{{\sin y}}{{\cos y}} = \frac{{4/5}}{{3/5}} = \frac{4}{3}$

…….(iii)
We know that,
$\tan (x + y) = \frac{{\tan x + \tan y}}{{1 - \tan x \cdot \tan y}}$

$\Rightarrow$ $\tan (x + y) = \frac{{\frac{5}{{12}} + \frac{4}{3}}}{{1 - \frac{5}{{12}} \cdot \frac{4}{3}}} \Rightarrow \tan (x + y) = \frac{{\frac{{15 + 48}}{{36}}}}{{\frac{{36 - 20}}{{36}}}}$

$\Rightarrow$ $\tan (x + y) = \frac{{63/36}}{{16/36}}$

$\Rightarrow$ $\tan (x + y) = \frac{{63}}{{16}}$
$\Rightarrow$ $x + y = {\tan ^{ - 1}}\frac{{63}}{{16}}$

$\Rightarrow$ ${\tan ^{ - 1}}\frac{5}{{12}} + {\tan ^{ - 1}}\frac{4}{3} = {\tan ^{ - 1}}\frac{{63}}{{16}}$ Hence proved