证明:∵tanβ=
sinβ
cosβ =
sinαcos(α+β)
cosβ =
sinα(cosαcosβ-sinαsinβ)
cosβ =sinαcosα-sin 2αtanβ
∴(1+sin 2α)tanβ=sinαcosα
∴tanβ=
sinαcosα
1+si n 2 α =
tanα
1+si n 2 α
co s 2 α =
tanα
co s 2 α+2si n 2 α
co s 2 α =
tanα
1+2ta n 2 α
(2)∵tanα>0,tanβ>0
∴tanβ=
1
1
tanα +2tanα ≤
1
2
2
当且仅当
1
tanα =2tanα ,即tanα=
2
2 时,
tanβ max=
2
2
1+2×
1
2 =
2
4
∴tan(α+β)=
2
2 +
2
4
1-
2
2 ×
2
4 =
3
2
4 ×
4
3 =
2