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

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

$LHS=\sin^{-1}\frac{8}{17}+\sin^{-1}\frac{3}{5}$

$let \: \: \sin^{-1}\frac{8}{17}=\theta_{1}$

$\Rightarrow \sin \theta_{1}=\frac{8}{17}$

$\Rightarrow \tan \theta_{1}=\frac{8}{15}\Rightarrow \theta_{1}=\tan^{-1}\frac{8}{15}$

And, $\sin \frac{3}{5}=\theta_{2}\Rightarrow \sin^{-1}\frac{3}{5}$

$\Rightarrow \tan \theta_{2}=\frac{3}{4}\Rightarrow \theta_{2}=\tan^{-1}\frac{3}{4}$

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

$=\tan^{-1}\left [ \frac{\frac{8}{15}+\frac{3}{4}}{1-\frac{8}{15}\times \frac{3}{4}} \right ]$ $\left [ Since , \tan^{-1}x+\tan^{-1} y=tan^{-1}\left ( \frac{x+y}{1-xy} \right ) \right ]$

$=\tan^{-1}\left [ \frac{\frac{77}{60}}{\frac{36}{60}} \right ]$

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

Let $=\theta _{3}=tan^{-1}\left ( \frac{77}{36} \right )\Rightarrow \tan \theta_{3}=\frac{77}{36}$

$\Rightarrow \sin \theta_{3}=\frac{77}{\sqrt{5929+1296}}=\frac{77}{85}$

$\therefore \theta _{3}=\sin^{-1}\left ( \frac{77}{85} \right )$

$= \sin^{-1}\left ( \frac{77}{85} \right )=RHS$

Hence proved