Question

please answer it....
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Answer 1

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