in any Trianlge ABC prove that a^2sin(B-C)/(sinB+sinC) +b^2sin(C-A)/(sinC+sinA) + c^2sin(A-B)/(sinA+sinB) =0 Share with your friends Share 17 Rishabh Mittal answered this We know that by sine rule :asin A = bsin B = csin CLet asin A = bsin B = csin C =kLHS=a2 sinB-Csin B+sin C+b2 sinC-Asin C+sin A+c2 sinA-Bsin A+sin B=k2sin2A sinB-Csin B+sin C+k2sin2B sinC-Asin C+sin A+k2sin2C sinA-Bsin A+sin B=k2sin A sinπ-B+C sin B-Csin B+sin C+k2sin B sinπ-C+A sinC-Asin C+sin A+k2sin C sinπ-A+B sinA-Bsin A+sin B=k2sin A sinB+C sin B-Csin B+sin C+k2sin B sinC+A sinC-Asin C+sin A+k2sin C sinA+B sinA-Bsin A+sin B=k2 sin Asin2B-sin2Csin B+sin C+k2 sin Bsin2C-sin2Asin C+sin A+k2 sin Csin2A-sin2Bsin A+sin B=k2 sin Asin B+sin Csin B-sin Csin B+sin C+k2 sin B sin C+sin Asin C-sin Asin C+sin A+k2 sin C sin A+sin Bsin A-sin Bsin A+sin B=k2sin Asin B-sin C+sin B sin C-sin A+sin C sin A-sin B=k2sin A sin B-sin A sin C+sin B sin C-sin B sin A+sin C sin A-sin C sin B=0=RHSHence Proved. 26 View Full Answer
