In a ABC, prove that
(ii) sin A sin (B-C) + sin B sin (C - A) + sin C sin
(A-B) = 0.
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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
Step-by-step explanation:
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