已知1−cos2αsinαcosα=1,tan(β−α)=−1/3,则tan(β-2α)等于_.

问题描述:

已知

1−cos2α
sinαcosα
=1,tan(β−α)=−
1
3
,则tan(β-2α)等于______.

1−cos2α
sinαcosα
=
1−(1−2sin2α)
sinαcosα
=2tanα=1,得到tanα=
1
2
,又tan(β−α)=−
1
3

则tan(β-2α)=tan[(β-α)-α]=
tan(β−α)−tanα
1+tan(β−α)tanα
=
1
3
1
2
1−
1
6
=-1.
故答案为:-1