please answer it....

Answers
Answer:
To Prove: SinA . Sin(B - C) + SinB . Sin(C - A) + SinC . Sin(A - B)=0
=>SinA . (SinBCosC - CosBSinC) + SinB . (SinCCosA - CosCSinA) + SinC . (SinACosB - CosASinB)
=>SinASinBCosC -SinACosBSinC + SinBSinCCosA - SinBCosCSinA + SinCSinACosB - SinCCosASinB
All get cancelled
=>0
Hence Proved
To Prove: cos A . sin ( B - C ) + cos B . sin ( C - A ) + cos C . sin ( A - B ) = 0
We use the the following result,
sin ( x - y ) = sin x . cos y - cos x . sin y
LHS
= cos A . sin ( B - C ) + cos B . sin ( C - A ) + cos C . sin ( A - B )
= cos A ( sin B . cos C - cos B . sin C ) + cos B ( sin C . cos A - cos C . sin A ) + cos C ( sin A . cos B - cos A . sin B )
= cos A . sin B . cos C - cos A . cos B . sin C + cos B . sin C . cos A - cos B . cos C . sin A + cos C . sin A . cos B - cos C . cos A . sin B
= 0 + 0 + 0
= 0
RHS
= 0
LHS = RHS
Hence, Proved