on simplification we have,
\[\begin{array}{*{35}{l}}
=\text{ }{{k}^{2}}~[sin\text{ }A\text{ }sin\text{ }\left( B\text{ }\text{ }C \right)\text{ }+\text{ }sin\text{ }B \\
sin\text{ }\left( C\text{ }\text{ }A \right)\text{ }+\text{ }sin\text{ }C\text{ }sin\text{ }\left( A\text{ }\text{ }B \right)] \\
\end{array}\]
We know,
\[~sin\text{ }\left( A\text{ }\text{ }B \right)\text{ }=\text{ }sin\text{ }A\text{ }cos\text{ }B\text{ }\text{ }cos\text{ }A\text{ }sin\text{ }B\]
\[Sin\text{ }\left( B\text{ }\text{ }C \right)\text{ }=\text{ }sin\text{ }B\text{ }cos\text{ }C\text{ }\text{ }cos\text{ }B\text{ }sin\text{ }C\]
\[Sin\text{ }\left( C\text{ }\text{ }A \right)\text{ }=\text{ }sin\text{ }C\text{ }cos\text{ }A\text{ }\text{ }cos\text{ }C\text{ }sin\text{ }A\]
So the above equation now becomes,
\[\begin{array}{*{35}{l}}
=\text{ }{{k}^{2}}~[sin\text{ }A\text{ }\left( sin\text{ }B\text{ }cos\text{ }C\text{ }\text{ }cos\text{ }B\text{ }sin\text{ }C \right)\text{ }+ \\
sin\text{ }B\text{ }\left( sin\text{ }C\text{ }cos\text{ }A\text{ }\text{ }cos\text{ }C\text{ }sin\text{ }A \right)\text{ }+\text{ }sin\text{ }C\text{ }\left( sin\text{ }A\text{ }cos\text{ }B\text{ }\text{ }cos\text{ }A\text{ }sin\text{ }B \right)] \\
\end{array}\]
\[\begin{array}{*{35}{l}}
=\text{ }{{k}^{2}}~[sin\text{ }A\text{ }sin\text{ }B\text{ }cos\text{ }C\text{ }\text{ }sin\text{ }A\text{ }cos\text{ }B\text{ }sin\text{ }C\text{ }+\text{ }sin\text{ }B\text{ }sin\text{ }C\text{ }cos\text{ }A\text{ }\text{ }sin\text{ }B \\
cos\text{ }C\text{ }sin\text{ }A\text{ }+\text{ }sin\text{ }C\text{ }sin\text{ }A\text{ }cos\text{ }B\text{ }\text{ }sin\text{ }C\text{ }cos\text{ }A\text{ }sin\text{ }B)] \\
\end{array}\]
on simplification we have,
\[=\text{ }0\] = RHS
Hence proved.