Sum of inverses is an inverse

Let $\alpha = \sin^ {-1} \frac {3}{5}, \beta = \sin^{-1} \frac {8}{17}$

Then sin $\alpha = \frac {3}{5}$ , cos $\alpha = \frac {4}{5}$

and sin $\beta = \frac {8}{17}$ , cos $\alpha = \frac {15}{17}$

Now $\sin ( \alpha + \beta) = sin \alpha \times cos \beta + cos \alpha \times sin \beta = \frac{3}{5} \times \frac {15}{17} + \frac{4}{5} \times \frac{8}{17} = \frac {45}{85} + \frac{32}{85} = \frac{77}{85}$

So $\sin ^{-1} \frac {3}{5} + \sin ^{-1} \frac {8}{17} = \sin ^{-1} \frac {77}{85}$

PS

It should $\alpha + \beta = \sin ^{-1} \frac {77}{85} + 2k \pi$

or

$\alpha + \beta =\pi - \sin ^{-1} \frac {77}{85} + 2k \pi$

but, since $\alpha, \beta < \pi /4$ we know that the above equation holds