Let us consider LHS:
\[si{{n}^{2}}~{{42}^{o}}~\text{ }co{{s}^{2}}~{{78}^{o}}~=\text{ }si{{n}^{2}}~({{90}^{o}}~\text{ }{{48}^{o}})\text{ }\text{ }co{{s}^{2}}~({{90}^{o}}~\text{ }{{12}^{o}})\]
\[=\text{ }co{{s}^{2}}~{{48}^{o}}~\text{ }si{{n}^{2}}~{{12}^{o}}~\left[ since,\text{ }sin\text{ }\left( 90\text{ }\text{ }A \right)\text{ }=\text{ }cos\text{ }A\text{ }and\text{ }cos\text{ }\left( 90\text{ }\text{ }A \right)\text{ }=\text{ }sin\text{ }A \right]\]
We know, \[cos\text{ }\left( A\text{ }+\text{ }B \right)\text{ }cos\text{ }\left( A\text{ }\text{ }B \right)\text{ }=\text{ }co{{s}^{2}}A\text{ }\text{ }si{{n}^{2}}B\]
Then the above equation becomes,
= \[co{{s}^{2}}~({{48}^{o}}~+\text{ }{{12}^{o}})\text{ }cos\text{ }({{48}^{o}}~\text{ }{{12}^{o}})\]
= \[cos\text{ }{{60}^{o}}~cos\text{ }{{36}^{o}}~[since,\text{ }cos\text{ }{{36}^{o}}~=\text{ }\left( \surd 5\text{ }+\text{ }1 \right)/4]\]
= \[1/2\text{ }\times \text{ }\left( \surd 5\text{ }+\text{ }1 \right)/4\]
\[=\text{ }\left( \surd 5\text{ }+\text{ }1 \right)/8\]
= RHS
Hence proved.