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

Answer

We have, $\sin^{-1}\frac{8}{17}+\sin^{-1}\frac{3}{5}=\sin^{-1}\frac{77}{85}$ $\therefore$ LHS $=\sin^{-1}\frac{8}{17}+\sin^{-1}\frac{3}{5}$ $=\sin^{-1}\Bigg(\frac{8}{17}\sqrt{1-\Big(\frac{3}{5}\Big)^2}+\frac{3}{5}\sqrt{1-\Big(\frac{8}{17}\Big)^2}\Bigg)$ $\Big[\because\ \sin^{-1}\text{x}+\sin^{-1}\text{y}=\sin^{-1}\Big(\text{x}\sqrt{1-\text{y}^2}+\text{y}\sqrt{1-\text{x}^2}\Big)\Big]$ $=\sin^{-1}\Big(\frac{8}{17}\times\frac{4}{5}+\frac{3}{5}\times\frac{15}{17}\Big)$$=\sin^{-1}\Big(\frac{32}{85}+\frac{45}{85}\Big)$
$=\sin^{-1}\Big(\frac{77}{85}\Big)$ = RHS

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