according to sine rule,
Now,
\[~c\text{ }=\text{ }k\text{ }sin\text{ }C\]
In the same way,
\[a\text{ }=\text{ }k\text{ }sin\text{ }A\]
also,
\[~b\text{ }=\text{ }k\text{ }sin\text{ }B\]
As we know,
\[let\text{ }LHS:\]
\[b\text{ }sin\text{ }B\text{ }\text{ }c\text{ }sin\text{ }C\]
Substituting the values of b and c,
\[k\text{ }sin\text{ }B\text{ }sin\text{ }B\text{ }\text{ }k\text{ }sin\text{ }C\text{ }sin\text{ }C\text{ }=\text{ }k\text{ }\left( si{{n}^{2}}~B\text{ }\text{ }si{{n}^{2}}~C \right)\text{ }\ldots \ldots \ldots .\left( i \right)\]
We know,
\[Si{{n}^{2}}~B\text{ }\text{ }si{{n}^{2}}~C\text{ }=\text{ }sin\text{ }\left( B\text{ }+\text{ }C \right)\text{ }sin\text{ }\left( B\text{ }\text{ }C \right),\]
Substituting the above values in equation (i),
\[k\text{ }\left( si{{n}^{2}}~B\text{ }\text{ }si{{n}^{2}}~C \right)\text{ }=\text{ }k\text{ }\left( sin\text{ }\left( B\text{ }+\text{ }C \right)\text{ }sin\text{ }\left( B\text{ }\text{ }C \right) \right)\]
\[[since,\text{ }\pi \text{ }=\text{ }A\text{ }+\text{ }B\text{ }+\text{ }C\Rightarrow ~B\text{ }+\text{ }C\text{ }=\text{ }\pi \text{ }A]\]
The above equation becomes,
\[=\text{ }k\text{ }\left( sin\text{ }\left( \pi \text{ }A \right)\text{ }sin\text{ }\left( B\text{ }\text{ }C \right) \right)\]
\[\left[ since,\text{ }sin\text{ }\left( \pi \text{ }\text{ }\theta \right)\text{ }=\text{ }sin\text{ }\theta \right]\]
\[=\text{ }k\text{ }\left( sin\text{ }\left( A \right)\text{ }sin\text{ }\left( B\text{ }\text{ }C \right) \right)\]
From sine rule, \[a\text{ }=\text{ }k\text{ }sin\text{ }A,\]
hence the above equation becomes,
\[=\text{ }a\text{ }sin\text{ }\left( B\text{ }\text{ }C \right)\]
\[=\text{ }RHS\]
Hence proved.