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已知5sin2α=sin2°,则tan(α+1°)tan(α−1°)的值是(  )

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已知5sin2α=sin2°,则
tan(α+1
已知5sin2α=sin2°,则tan(α+1°)tan(α−1°)的值是(  )
5sin2a=sin2°
5sin[(a+1°)+(a-1°)]
=sin[(a+1°)-(a-1°)]
=5sin(a+1°)cos(a-1°)+5cos(a+1°)sin(a-1°)
=sin(a+1°)cos(a-1°)-cos(a+1°)sin(a-1°)
∴4sin(a+1°)cos(a-1°)=-6cos(a+1°)sin(a-1°)
两边除以cos(a-1°)cos(a+1°):
得4tan(a+1°)=-6tan(a-1°)

tan(a+1°)
tan(a−1°)=-
6
4=-
3
2
故选:B.