已知5sin2a=sin2°，则tan(a+1°)/tan(a-1°)的值是多少？

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)：
4tan(a+1°)=-6tan(a-1°）

[tan(a+1°)]/[tan(a-1°)]=-6/4=-3/2.

解:
由于:5sin2a=sin2
则有:
5sin[(a+1)+(a-1)]=sin[(a+1)-(a-1)]
则有：
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，得：
4sin(a+1)cos(a-1)=-6cos(a+1)sin(a-1)

两边同时除以cos(a-1)cos(a+1),得：
4tan(a+1)=-6tan(a-1)
则：
tan(a+1)/tan(a-1)
=-3/2