\[sin\text{ }50{}^\circ \text{ }\text{ }sin\text{ }70{}^\circ \text{ }+\text{ }sin\text{ }10{}^\circ \text{ }=\text{ }0\]
Take LHS:
\[sin\text{ }50{}^\circ \text{ }\text{ }sin\text{ }70{}^\circ \text{ }+\text{ }sin\text{ }10{}^\circ \]
We know that:
\[sin\text{ }A\text{ }\text{ }sin\text{ }B\text{ }=\text{ }2\text{ }cos\text{ }\left( A+B \right)/2\text{ }sin\text{ }\left( A-B \right)/2\]
\[sin\text{ }50{}^\circ \text{ }\text{ }sin\text{ }70{}^\circ \text{ }+\text{ }sin\text{ }10{}^\circ \text{ }=\text{ }2\text{ }cos\text{ }({{50}^{o}}~+\text{ }{{70}^{o}})/2\text{ }sin\text{ }({{50}^{o}}~\text{ }{{70}^{o}})\text{ }+\text{ }sin\text{ }{{10}^{o}}\]
\[=\text{ }2\text{ }cos\text{ }{{120}^{o}}/2\text{ }sin\text{ }(-{{20}^{o}})/2\text{ }+\text{ }sin\text{ }{{10}^{o}}\]
\[=\text{ }2\text{ }cos\text{ }{{60}^{o}}~(-\text{ }sin\text{ }{{10}^{o}})\text{ }+\text{ }sin\text{ }{{10}^{o}}~\left[ since,\{sin\text{ }\left( -A \right)\text{ }=\text{ }-sin\text{ }\left( A \right)\} \right]\]
\[=\text{ }2\text{ }\times \text{ }1/2\text{ }\times \text{ }\text{ }sin\text{ }{{10}^{o}}~+\text{ }sin\text{ }{{10}^{o}}\]
\[=\text{ }0\]
= RHS
Hence proved.