在三角形ABC中 若a^2+b^2=2c^2 ,求sin^2A+sin^2B/sin^2C

a^2+b^2=2c^2
(a^2+b^2)/c^2=2
a/sinA=b/sinB=c/sinC
令1/k=a/sinA=b/sinB=c/sinC
所以sinA=ak,sinB=bk,sinC=ck
所以sin^2A+sin^2B/sin^2C
=(a^2k^2+b^2k^2)/c^2k^2
=(a^2+b^2)/c^2
=2