Question

In any triangle abc prove that a sin (b-c) +b sin (c-a) + c sin (a-b) =0

Answer 1

Answer:

sin(B-C)+b sin(C-A)+c sin(A-B)=0. ... Sin(C - A) + SinC . Sin(A - B)=0 ... =>0. Hence Proved.

Answer 2

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In any triangle ABC,

a/sin A = b/sin B = c/sin C = k

a = k sin A, b = k sin B, c = k sin C

LHS

= a sin (B – C) + b sin (C – A) + c sin (A – B)

= k sin A [sin B cos C – cos B sin C] + k sin B [sin C cos A – cos C sin A] + k sin C [sin A cos B – cos A sin B]

= k sin A sin B cos C – k sin A cos B sin C + k sin B sin C cos A – k sin B cos C sin A + k sin C sin A cos B – k sin C cos A sin B

= 0

= RHS

Hence proved that a sin (B – C) + b sin (C – A) + c sin (A – B) = 0

Hope it's Helpful....:)