In any triangle ABC prove that sinA+sinB+sinC-sin(A+B+C)= 4sin(A+B/2)sin(B+C/2)sin(C+A/2)

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`(a^2sin(B-C))/(sinB+sinC)+(b^2sin(C-A))/(sinC+sinA)+(c^2sin(A-B))/(sinA+sinB=0)``(a^2sin(B-C))/(sinB+sinC)+(b^2sin(C-A))/(sinC+sinA)+(c^2sin(A-B))/(sinA+sinB=0)`