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In triangle ABC a²sin2B - b²sin2A then find the value of it
a²sin2B + b²sin2A = c²cotC/2 4R².sin²A.sin2B + 4R².sin²B.sin2A = 4R².sin²C.cotC/2 sin²A.sin2B + sin²B.sin2A = sin²C.cotC/2 (1 - cos2A)/2.sin2B + (1 - cos2B)/2.sin2A = sin²C.cotC/2 sin2B - cos2A.sin2B + sin2A - cos2B.sin2A = 2sin²C.cotC/2 sin2A + sin2B - sin(2A + 2B) = 2sin²C.cotC/2 ( vì sin2A.cos2B + cos2A.sin2B = sin (2A +2B) sin2A + sin2B + sin2C = 2sin²C.cotC/2 2sin(A + B)cos(A - B) + 2sinC.cosC = 2sin²C.cotC/2 2sinC.[ cos(A - B) - cos(A + B) ] = 2sin²C.cotC/2 2sinC.-2.sinA.sin(-B) = 2sin²C.cotC/2 4sinA.sinB.sinC = 2sin²C.cotC/2 2sinA.sinB = sinC.cotC/2 2sinA.sinB = 2sinC/2.cosC/2.cosC/2 /sinC/2 sinA.sinB = cos²C/2 2sinA.sinB = 1 + cosC 2sinA.sinB = 1 - cos(A + B) 2sinA.sinB = 1 - cosA.cosB + sinA.sinB cosA.cosB + sinA.sinB = 1 cos(A - B) = 1 A = B A - B =0
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