问题 解答题
已知tan(α-β)=
sin2β
5-cos2β
,α、β≠kπ+
π
2
,kϵZ,求证:2tanα=3tanβ.
答案

证明:∵tan(α-β)=

tanα-tanβ
1+tanαtanβ

sin2β
5-cos2β
=
2sinβcosβ
5-(1-2sin2β)
=
2sinβcosβ
4+2sin2β
=
sinβcosβ
2+sin2β

=

sinβcosβ
2cos2β+3sin2β
=
tanβ
2+3tan2β
=
3
2
tanβ-tanβ
1+
3
2
tan2β

且tan(α-β)=

sin2β
5-cos2β

tanα-tanβ
1+tanαtanβ
=
3
2
tanβ-tanβ
1+
3
2
tan2β

则tanα=

3
2
tanβ,即2tanα=3tanβ.

判断题
判断题