已知 5sin2a=sin2度 则 tan(a +1度）/ tan(a -1度） =

∵5sin2a=sin2°
∴4sin2a+sin2a=sin2°
sin2a-sin2°=-4sin2a
2cos[（2a+2°）/2]sin[(2a-2°)/2]=-4sin2a
cos(a+1°)sin(a-1°)=-2sin2a.1式
又∵5sin2a+sin2a=sin2°+sin2a
∴6sin2a=2sin[(2a+2°)/2]cos[(2a-2°)/2]
3sin2a=sin(a+1°)cos(a-1°).2式
tan（a+1°）/tan(a-1°)
=[sin(a+1°)/cos(a+1°)]/[sin(a-1°)/cos(a-1°)]
=[sin(a+1°)cos(a-1°）]/[cos(a+1°)sin(a-1°)].根据1式2式结果
=3sin2a/(-2sin2a)
=-3/2

5sin[(a+1)+(a-1)]=sin[(a+1)-(a-1)]
展开5sin(a+1)cos(a-1)+5cos(a+1)sin(a-1)=sin(a+1)cos(a-1)-cos(a+1)sin(a-1)
整理，4sin(a+1)cos(a-1）=-6cos(a+1)sin(a-1)
最后结果，-3/2