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

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sinacosb+sinbcosa=3/5
sinacosb-sinbcosa=-1/5
sinacosb=1/5
sinbcosa=2/5
sinacosb/sinbcosa=tana/tanb=1/2
C是锐角,所以A+B是钝角
cos(A+B)=-4/5,cos(A-B)=2√6/5
cosAcosB-sinAsinB=-4/5
cosAcosB+sinAsinB=2√6/5
sinAsinB=(2+√6)/5
三角形面积=absinC/2=AB*h/2
sinC=sin(A+B)=3/5
ab*3/5=3*h
h=ab/5
因为c/sinC=3/(3/5)=5
所以a/sinA=b/sinB=5
a=5sinA,b=5sinB
ab=25sinAsinB
h=ab/5
=5sinAsinB=2+√6