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

Solving LHS, $\sin^{-1}\frac{5}{13}+\cos^{-1}\frac{3}{5}$

$Let \: \sin^{-1}\frac{5}{13}=x$

$\Rightarrow \sin x=\frac{5}{13}$

$And \, \cos^{2}x=1-\sin^{2}x$

$\Rightarrow 1-\frac{25}{169}=\frac{144}{169}$

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

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

$\Rightarrow \tan x=\frac{5}{12}..........(i)$

Again , let $\cos^{-1}\frac{3}{5}=y$

$\Rightarrow \cos y=\frac{3}{5}$

$\therefore \sin y=\sqrt{1-\cos^{2}y}$

$\Rightarrow \sin y=\sqrt{1-\left (\frac{3}{5} \right )^{2}}$

$\Rightarrow \sin y=\sqrt{\frac{16}{25}}=\frac{4}{5}$

$\Rightarrow \tan y=\frac{\sin y}{\cos y}=\frac{\frac{4}{5}}{\frac{3}{5}}=\frac{4}{3}.........(ii)$

We know that,  $\tan\left ( x+y \right )=\frac{\frac{5}{12}+\frac{4}{3}}{1-\frac{5}{12}\frac{4}{3}}$  [from (i),(ii)]

$\Rightarrow \tan\left ( x+y \right )=\frac{\frac{15+48}{36}}{\frac{36-20}{36}}$

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

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

Since , LHS=RHS

Hence Proved.