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If A+B+C=0 Then prove that;
sin(B+C-A)+sin(C+A-B)+sin(A+B-C)=4sinA.sinB.sinC

Since in a triangle A + B + C = π,
=> sin( B+C-A ) + sin( C+A-B ) + sin( A+B-C )
=> sin( π - 2A ) + sin( π - 2B ) + sin( π - 2C )
=> sin (2A ) + sin( 2B ) + sin( 2C )
=> 2sin( A + B)*cos(A - B) + 2sin( C)*Cos(C) ; use sinC + sinD = 2sin(C+D)/2 * cos(C -D)/2,
=> 2sin( π - C)*cos(A - B) + 2sin( C)* cos(C)
=> 2sin(C)*cos(A - B) + 2sin( C)*cos{π - (A+B)} ; use A + B + C = π
=> 2sin(C)[cos(A - B) - Cos(A+B)] ; use cos (π - θ) = -cosθ
=> 2sin(C)[2sin(A) *sin(B)] ; use cos C - CosD = 2sin(C+D)/2 * sin(D - C)/2
=> 4sin(A) sin(B) sin(C) ----- RHS hence Proved

Hope this helped

A+B+C = 180 so A+B= 180-C, B+C= 180-A, A+C= 180 -B
sin(B+C -A) = sin( 180- 2A)= sin2A
sin( C+A-B) = sin(180 - 2B) = sin2B
sin(A+B-C) = sin( 180- 2C) = sin2C
so L.H.S,= sin(B+C-A)+ sin(C+A-B)+sin(A+B-C)= sin2A +sin2B + sin2C
= 2sin(A+B) cos(A- B) + sin2C
= 2sin( 180- C) cos(A-B)+ sin2C
= 2sinC cos( A- B) +2sinC cosC
= 2sinC[ cos(A-B) + cosC]
=2 sinC* [ cos { A+C -B}/2 cos{A -(B+C)}/2 ]
= 2 sinC * 2cos(180-2B)/2 cos{ A - (180-A)/2}
= 4sinC * cos(90- B) * cos-( 90- A)
= 4sinC sinB * * cos( 90- A) { cos (-theta) = costheta }
= 4 sinA sinB sinC = R.H.S.
======================================...
sin(B+C-A)+sin(C+A-B)+sin(A+B-C)
=2sin{B+C-A+C+A-B}/2 cos{B+C-A-C-A+B}/2 + sin(A+B-C)
= 2sinC cos (B-A) + sin(A+B-C)
= 2 sinC [ cos(B-A) + sin{ 180 -2C}]
= 2sinC { cos(B-A) + sin2C }
= 2sinC [ cos(A-B) + 2sinC cosC]
= 2sinC { cos(A-B) + cosC ]
=2 sinC { 2 cos(A-B+C)/2 cos(A-B-C)/2}
= 4 sinC * cos{- B +(180- B)}/2 cos { A- (180-A)}/2
= 4 sinC cos(90-B) cos -(90- A)
= 4 sinC sinB sinA = R.H.S. [ cos{ -theta = cos theta ]