Considering L.H.S.
\[c\text{ }\left( a\text{ }cos\text{ }B\text{ }\text{ }b\text{ }cos\text{ }A \right)\]
\[ca\text{ }cos\text{ }B\text{ }and\text{ }cb\text{ }cos\text{ }A\] is present in L.H.S, which can be obtained from cosine formulae.
Using cosine formula:
\[Cos\text{ }A\text{ }=\text{ }({{b}^{2}}~+\text{ }{{c}^{2}}~\text{ }{{a}^{2}})/2bc\]
\[bc\text{ }cos\text{ }A\text{ }=\text{ }({{b}^{2}}~+\text{ }{{c}^{2}}~\text{ }{{a}^{2}})/2\text{ }\ldots \text{ }\left( i \right)\]
Or,
\[Cos\text{ }B\text{ }=\text{ }({{a}^{2}}~+\text{ }{{c}^{2}}~\text{ }{{b}^{2}})/2ac\]
\[ac\text{ }cos\text{ }B\text{ }=\text{ }({{a}^{2}}~+\text{ }{{c}^{2}}~\text{ }{{b}^{2}})/2\text{ }\ldots \text{ }\left( ii \right)\]
Subtracting equation (ii) from (i) we get,
\[ac\text{ }cos\text{ }B\text{ }\text{ }bc\text{ }cos\text{ }A\]
\[=\text{ }({{a}^{2}}~+\text{ }{{c}^{2}}~\text{ }{{b}^{2}})/2\text{ }\text{ }({{b}^{2}}~+\text{ }{{c}^{2}}~\text{ }{{a}^{2}})/2\]
So,
\[=\text{ }{{a}^{2}}~\text{ }{{b}^{2}}\]
\[\therefore c\text{ }\left( a\text{ }cos\text{ }B\text{ }\text{ }b\text{ }cos\text{ }A \right)\text{ }=\text{ }{{a}^{2}}~\text{ }{{b}^{2}}\]
Hence proved.