Prove that $\sin ^{-1} \frac{8}{17}+\sin ^{-1} \frac{3}{5}=\tan ^{-1} \frac{77}{36}$

Let $\sin ^{-1} \frac{8}{17}=x$. Then,$\sin x=\frac{8}{17}$.
Since $\cos x=\sqrt{1-\sin^2 x}=\sqrt{1-(\frac{8}{17})^2}=\sqrt{\frac{289-64}{289}}=\frac{15}{17}$,we have $\tan x=\frac{\sin x}{\cos x}=\frac{8}{15}$.
Thus,$\sin ^{-1} \frac{8}{17}=\tan ^{-1} \frac{8}{15}$ $...(1)$
Let $\sin ^{-1} \frac{3}{5}=y$. Then,$\sin y=\frac{3}{5}$.
Since $\cos y=\sqrt{1-\sin^2 y}=\sqrt{1-(\frac{3}{5})^2}=\sqrt{\frac{25-9}{25}}=\frac{4}{5}$,we have $\tan y=\frac{\sin y}{\cos y}=\frac{3}{4}$.
Thus,$\sin ^{-1} \frac{3}{5}=\tan ^{-1} \frac{3}{4}$ $...(2)$
Now,taking the $L.H.S.$:
$\sin ^{-1} \frac{8}{17}+\sin ^{-1} \frac{3}{5} = \tan ^{-1} \frac{8}{15}+\tan ^{-1} \frac{3}{4}$
Using the formula $\tan ^{-1} a + \tan ^{-1} b = \tan ^{-1} \left(\frac{a+b}{1-ab}\right)$:
$= \tan ^{-1} \left(\frac{\frac{8}{15}+\frac{3}{4}}{1-\frac{8}{15} \times \frac{3}{4}}\right)$
$= \tan ^{-1} \left(\frac{\frac{32+45}{60}}{\frac{60-24}{60}}\right)$
$= \tan ^{-1} \left(\frac{77}{36}\right) = R.H.S.$