Question

In triangle ABC, prove that a sin (B-C)+b sin (C-A)+c sin (A-B)=0​

Answer 1

               $\huge\boxed{Answer:}$

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

                    $\huge\boxed{ Proved}$

Answer 2

Answer:

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