Show that $\sin ^{-1} \frac{3}{5}-\sin ^{-1} \frac{8}{17}=\cos ^{-1} \frac{84}{85}$

Let $\sin ^{-1} \frac{3}{5} = x$ and $\sin ^{-1} \frac{8}{17} = y$.
Then,$\sin x = \frac{3}{5}$ and $\sin y = \frac{8}{17}$.
Since $\cos x = \sqrt{1 - \sin^2 x}$,we have $\cos x = \sqrt{1 - (\frac{3}{5})^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$.
Similarly,$\cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - (\frac{8}{17})^2} = \sqrt{1 - \frac{64}{289}} = \sqrt{\frac{225}{289}} = \frac{15}{17}$.
Using the identity $\cos(x - y) = \cos x \cos y + \sin x \sin y$,we get:
$\cos(x - y) = (\frac{4}{5})(\frac{15}{17}) + (\frac{3}{5})(\frac{8}{17}) = \frac{60}{85} + \frac{24}{85} = \frac{84}{85}$.
Thus,$x - y = \cos^{-1} \frac{84}{85}$.
Substituting back $x$ and $y$,we get $\sin^{-1} \frac{3}{5} - \sin^{-1} \frac{8}{17} = \cos^{-1} \frac{84}{85}$.