(1)sin(A+B)=
,sin(A-B)=
sin(A+B)=sinAcosB+sinBcosA=
sin(A- B)=sinAcosB-sinBcosA=
两式相加相减后可得:sinAcosB=
,sinBcosA=
将两式相除,可得tanA=2tanB
(2)∵△ABC是锐角三角形
∴0<C<
又A+B=π-C
∴
<A+B<π
∵sin(A+B)=3/5
∴cos(A+B)=
=-
则tan(A+B)=sin(A+B)/cos(A+B)=-
即(tanA+tanB)/(1-tanAtanB)=-
又tanA=2tanB
∴3tanB/(1-2tan²B)=-
即2tan²B-4tanB-1=0
解得tanB=
∵0<B<
∴tanB=
=1+