SOLUTION : $\quad \because \quad \frac{3}{5}>0, \frac{15}{17}>0$ and $\left(\frac{3}{5}\right)^2+\left(\frac{15}{17}\right)^2=\frac{8226}{7225}>1$
$\therefore \quad \sin ^{-1} \frac{3}{5}+\sin ^{-1} \frac{15}{17} =\pi-\sin ^{-1}\left(\frac{3}{5} \sqrt{1-\frac{225}{289}}+\frac{15}{17} \sqrt{1-\frac{9}{25}}\right) \\$
$=\pi-\sin ^{-1}\left(\frac{3}{5} \cdot \frac{8}{17}+\frac{15}{17} \cdot \frac{4}{5}\right)=\pi-\sin ^{-1}\left(\frac{84}{85}\right)$
