Let $\sin^{-1}\left(\frac{8}{17}\right) = \theta_1$ and $\sin^{-1}\left(\frac{3}{5}\right) = \theta_2$

Then $\sin \theta_1 = \frac{8}{17}$ and $\sin \theta_2 = \frac{3}{5}$

Finding $\cos \theta_1$ and $\cos \theta_2$:

For $\theta_1$:

$\cos^2 \theta_1 = 1 - \sin^2 \theta_1 = 1 - \left(\frac{8}{17}\right)^2 = 1 - \frac{64}{289} = \frac{225}{289}$

Therefore, $\cos \theta_1 = \frac{15}{17}$

For $\theta_2$:

$\cos^2 \theta_2 = 1 - \sin^2 \theta_2 = 1 - \left(\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{16}{25}$

Therefore, $\cos \theta_2 = \frac{4}{5}$

Finding $\tan \theta_1$ and $\tan \theta_2$:

$\tan \theta_1 = \frac{\sin \theta_1}{\cos \theta_1} = \frac{\frac{8}{17}}{\frac{15}{17}} = \frac{8}{15}$

$\tan \theta_2 = \frac{\sin \theta_2}{\cos \theta_2} = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4}$

Using the formula for $\tan(\theta_1 + \theta_2)$:

$\tan(\theta_1 + \theta_2) = \frac{\tan \theta_1 + \tan \theta_2}{1 - \tan \theta_1 \cdot \tan \theta_2}$

$= \frac{\frac{8}{15} + \frac{3}{4}}{1 - \frac{8}{15} \times \frac{3}{4}}$

$= \frac{\frac{32}{60} + \frac{45}{60}}{1 - \frac{24}{60}}$

$= \frac{\frac{77}{60}}{\frac{36}{60}}$

$= \frac{77}{36}$

Therefore, $\theta_1 + \theta_2 = \tan^{-1}\left(\frac{77}{36}\right)$

Hence proved:

$\sin^{-1}\left(\frac{8}{17}\right) + \sin^{-1}\left(\frac{3}{5}\right) = \tan^{-1}\left(\frac{77}{36}\right)$