Solution

Let \[ \sin^{-1}\left(\frac{8}{17}\right) = \theta_1, \quad \sin^{-1}\left(\frac{3}{5}\right) = \theta_2 \]

Then, \[ \tan \theta_1 = \frac{8}{15}, \quad \tan \theta_2 = \frac{3}{4} \]

Therefore, \[ \theta_1 + \theta_2 = \tan^{-1}\left(\frac{8}{15}\right) + \tan^{-1}\left(\frac{3}{4}\right) \]

\[ = \tan^{-1}\left( \frac{\frac{8}{15} + \frac{3}{4}} {1 – \frac{8}{15}\cdot\frac{3}{4}} \right) \]

\[ = \tan^{-1}\left(\frac{77}{36}\right) \]

Let \[ \theta_3 = \tan^{-1}\left(\frac{77}{36}\right) \]

Then, \[ \sin \theta_3 = \frac{77}{\sqrt{77^2 + 36^2}} = \frac{77}{85} \]

\[ \boxed{ \sin^{-1}\left(\frac{8}{17}\right) + \sin^{-1}\left(\frac{3}{5}\right) = \sin^{-1}\left(\frac{77}{85}\right) } \]

Alternate Method:

Let \[ \sin^{-1}\left(\frac{8}{17}\right) = x, \quad \sin^{-1}\left(\frac{3}{5}\right) = y \]

\[ \sin x = \frac{8}{17}, \quad \cos x = \frac{15}{17} \]

\[ \sin y = \frac{3}{5}, \quad \cos y = \frac{4}{5} \]

Using \[ \sin(x+y) = \sin x \cos y + \cos x \sin y \]

\[ = \frac{8}{17}\cdot\frac{4}{5} + \frac{15}{17}\cdot\frac{3}{5} = \frac{77}{85} \]

\[ \boxed{ \sin^{-1}\left(\frac{8}{17}\right) + \sin^{-1}\left(\frac{3}{5}\right) = \sin^{-1}\left(\frac{77}{85}\right) } \]