化简(1+cos2β+cos2β)/(1-cos2β+sin2β)+(1-cos2β-sin2β）/1+cos2β-sin2β）=

题目是(1+cos2β+sin2β)/(1-cos2β+sin2β)+(1-cos2β-sin2β）/1+cos2β-sin2β）吧?
=(1+cos²β-sin²β+2sinβcosβ)/(1-cos²β+sin²β+2sinβcosβ)+(1-cos²β+sin²β-2sinβcosβ)/(1+cos²β-sin²β-2sinβcosβ)
=(2cos²β+2sinβcosβ)/(2sin²β+2sinβcosβ)+(2sin²β-2sinβcosβ)/(2cos²β-2sinβcosβ)
=cosβ(cosβ+sinβ)/[sinβ(sinβ+cosβ)]+sinβ(sinβ-cosβ)/[cosβ(cosβ-sinβ)]
=cosβ/sinβ-sinβ/cosβ
=(cos²β-sin²β)/sinβcosβ
=2cos2β/sin2β