已知锐角△ABC中,sin(A+B)=3/5,sin(A-B)=1/5求tanB

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
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