tan(a+b)=2tan(a-b),则sin2b/sin2a=

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tan(a+b)=2tan(a-b),则sin2b/sin2a=
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tan(a+b)=2tan(a-b),则sin2b/sin2a=
tan(a+b)=2tan(a-b),则sin2b/sin2a=

tan(a+b)=2tan(a-b),则sin2b/sin2a=
我认为应该是这样算的tan(a+b)=2tan(a-b)
tan(a-b) /tan(a+b) =1/2
sin2b/tan2a
=[sin(b+a)cos(b-a)+cos(b+a)sin(b-a)]/[sin(b+a)cos(b-a)-cos(b+a)sin(b-a)]
=[1+tan(b-a)/tan(b+a)]/[1-tan(b-a)/tan(b+a)]
=(1-1/2)/(1+1/2)
=1/3