α、β∈（0，∏/2）,且α+β≠∏/2,α和β满足sinβ=sinαcos(α+β).用tanα表示tanβ；求tanβ的最大值

sinβ=sinαcos(α+β)=sinacosβcosa-sinasinasinβ
tanβ(1+sin^2a)=sinacosa=1/2sin2a
tanβ=1/2sin2a/[1+(1-cos2a)/2]=sin2a/(3-cos2a)
sin2a=2tana/(1+tan^2a),cos2a=(1-tan^2a)/(1+tan^2a)代入
得出：tanβ=tana/(1+2tan^2a)=1/(1/tana+2tana)
1/tana+2tana>=2根号2
所以tanβ<=4分之根号2

1)
sinβ=sinαcos(α+β).
=sinacosβcosa-sinasinasinβ
tanβ(1+sin^2a)=sinacosa
tanβ=sinacosa/(2sina^2+cos^2a)
=tana/(2tana+1/tana)
=tan^2a/(2tan^2a+1)
2)
tanβ=tan^2a/(2tan^2a+1)
=1/2（2tan^2a+1-1）/(2tan^2a+1)
=1/2-1/2*1/(2tan^2a+1)

tanβ<1/2
没有最大值

sinβ=sinαcos(α+β).
=sinacosβcosa-sinasinasinβ
tanβ(1+sin^2a)=sinacosa
tanβ=sinacosa/(2sina^2+cos^2a)
=tana/(2tana+1/tana)
=tan^2a/(2tan^2a+1)

tanβ=tan^2a/(2tan^2a+1)
=1/2（2tan^2a+1-1）/(2tan^2a2tan^2a+1)

tanβ<1/2有最大值
好难