已知锐角三角形ABC中，sin(A+B)=3/5,sin(A-B)=1/5.(1)求证:tanA=2tanB

（1）sin(A+B)=3/5,sin(A-B)=1/5
sin(a+b)=sinAcosB+sinBcosA=3/5
sin(a-b)=sinAcosB-sinBcosA=1/5
两式相加相减后可得：
sinAcosB=2/5
sinBcosA=1/5
将两式相除，可得tanA=2tanB
（2）tan(B)=sinB/cosB=sinBcosA/cosAcosB
cos(A)cos(B)=1/2[cos(A+B)+cos(A-B)]=1/2[4/5+2根号6/5]=(根号6-2)/5
tanB=1/(根号6-2)=(根号6+2)/2

(1)
tan(A)/tan(B)
=(sin(A)/cos(A))/(sin(B)/cos(B))
=sin(A)cos(B)/(sin(B)cos(A))

sin(A)cos(B)=1/2*[sin(A+B)+sin(A-B)]=2/5
sin(B)cos(A)=1/2*[sin(A+B)-sin(A-B)]=1/5

代入即得证

(2)
tan(B)=sin(A)/cos(A)=sin(A)cos(B)/cos(A)cos(B)
cos(A)cos(B)=1/2[cos(A+B)+cos(A-B)]=1/2[4/5+2根号6/5]=(2+根号6)/5

解：
1、因为sin(A+B)=3/5,sin(A-B)=1/5
则sin(A+B)=sinAcosB+cosAsinB=3/5 ①
sin(A-B)=sinAcosB-cosAsinB=1/5 ②
①+ ②得：sinAcosB=2/5

①- ②得：cosAsinB=1/5

则两式一比可得：sinAcosB/cosAsinB=2

即tanA=2tanB

2、

sinacosb+sinbcosa=3/5
sinacosb-sinbcosa=1/5
sinacosb=2/