1.

In any ΔABC, prove thatasin(B - C) + bsin(C – A) + csin(A - B) = 0

Answer»

LHS = ksinAsin(B - C) + ksinBsin(C – A) + ksinCsin(A – B) 

= k[sin(B + C)sin(B – C)+sin(C + A)sin(C – A) + sin(A + B) sin(A – B)] 

= K[sin2B – sin2C + sin2  C -sin2A + sin2A – sin2B] 

= k x 0 =0

= RHS.



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