In a triangle ABC, if a, b and c are the lengths of the sides opposite to the angles A, B and C respectively, then what is sin(A-B)/sin(A+B) equal to?

sin(A+B) = sin(π - C) = sin C. Using projection: sin(A-B)/sin(A+B) = sin(A-B)/sin C. Multiplying by sin C/sin C in expanded form using sine rule (a = k sin A, b = k sin B, c = k sin C): sin(A-B) sin(A+B) = sin²A - sin²B = (a² - b²)/k². And sin²C = c²/k². So sin(A-B)/sin C = (sin²A - sin²B)/(sin C · sin(A+B)) · sin(A+B). Direct: sin(A-B)/sin(A+B) = (sin²A - sin²B)/sin²(A+B) · sin(A+B)/sin(A-B)... Simpler: sin(A-B)/sin(A+B) = (a²-b²)/c².