设a.b.c依次是△ABC中角A.B.C所对的边,若tanAtanB/（tanA+tanB）=1005tanC,且a^2+b^2=mc^2,则实数m的值为

tanAtanB/[（tanA+tanB）tanC] =1005,
tanAtanB/[（tanA+tanB）tanC]
=sinAsinB/((sinAcosB+cosAsinB)tanC)
=sinAsinB/(tanCsin(A+B))
=sinAsinBcosC/(sinCsinC)
根据正弦定理得：sinA/sinC=a/c,sinB/sinC=b/c,
根据余弦定理得：cosC=(a^2+b^2-c^2)/2ab,
所以上式=ab/c^2*(a^2+b^2-c^2)/2ab
=1/2*(a^2+b^2-c^2)/ c^2
=1/2*(m-1),
根据已知：1/2*(m-1) =1005,m=2011.