是在△ABC中吧?
由正弦定理得：
a/c=sinA/sinC,b/c=sinB/sinC
(a²-b²)/c²
=(a/c)²-(b/c)²
=(sinA/sinC)²-(sinB/sinC)²
=(sinA+sinB)(sinA-sinB)/sin²C
=2cos[(A+B)/2]sin[(A-B)/2]*2cos[(A+B)/2]cos[(A-B)/2]/sin²C
=2cos²[(A+B)/2]*2sin[(A-B)/2]cos[(A-B)/2]/sin²C
=[cos(A+B)+1]*sin(A-B)/sin²C
=(1-cosC)*sin(A-B)/sin²C
只有当C=90°,原式成立