已知锐角三角形ABC中,sin(A+B)=3/5,sin(A-B)=1/5.(1证明tanA=2tanB(2)设AB=3,求AB边上的高

(1)sin(A＋B)/sin(A-B)=(sinAcosB+sinBcosA)/(sinAcosB-sinBcosA)=3,
(tanA+tanB)/(tanA-tanB)=3,
tanA+tanB=3(tanA-tanB)
tanA=2tanB
(2)cos(A+B)=4/5,cos(A-B)=2√6/5.
sinAsinB=(cos(A-B)-cos(A+B))/2=(2-√6)/5
h/tanA+h/tanB=3
h=3(tanA*tanB)/(tanA+tanB)
=3sinAsinB/sin(A+B)=2-√6