Let us consider LHS:
\[si{{n}^{2}}~{{24}^{o}}~\text{ }si{{n}^{2}}~{{6}^{o}}\]
we know, \[sin\text{ }\left( A\text{ }+\text{ }B \right)\text{ }sin\text{ }\left( A\text{ }\text{ }B \right)\text{ }=\text{ }si{{n}^{2}}A\text{ }\text{ }si{{n}^{2}}B\]
Then the above equation becomes,
\[si{{n}^{2}}~{{24}^{o}}~\text{ }si{{n}^{2}}~{{6}^{o}}~=\text{ }sin\text{ }({{24}^{o}}~+\text{ }{{6}^{o}})\text{ }\text{ }sin\text{ }({{24}^{o}}~\text{ }{{6}^{o}})\]
\[=\text{ }sin\text{ }{{30}^{o}}~\text{ }sin\text{ }{{18}^{o}}\]
\[=\text{ }sin\text{ }{{30}^{o}}~\text{ }\left( \surd 5\text{ }\text{ }1 \right)/4\text{ }[since,\text{ }sin\text{ }{{18}^{o}}~=\text{ }\left( \surd 5\text{ }\text{ }1 \right)/4]\]
\[=\text{ }1/2\text{ }\times \text{ }\left( \surd 5\text{ }\text{ }1 \right)/4\]
\[=\text{ }\left( \surd 5\text{ }\text{ }1 \right)/8\]
= RHS
Hence proved.