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.