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If any triangle ABC, find the value of `asin(B-C)+b sin(C-A)+c sin(A-B)dot`

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LHS = `k sin A sin(B - C) + k sin B sin(C – A) + k sin C sin (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 xx 0 =0` =
RHS.
Hence Prove
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