數學題[中3]` 急{20點}

1) 2tanθ/ (1+tan2次θ)
= 2tanθ/ [1+ (sin2次θ/cos2次θ)]
= 2tanθ/ [(cos2次θ+sin2次θ)/cos2次θ]
= 2tanθ/ (1/cos2次θ)
= 2tanθcos2次θ
= 2sinθcosθ

2) cos2次2θ+sin2次2θ= 1
3) sin2次θtanθ+cos2θ/tanθ+2sinθcosθ-1/tanθ
= sin2次θtanθ+ (cos2θ-1)/tanθ+2sinθcosθ
= sin2次θtanθ+ (1-2sin 2次θ-1)/tanθ+2sinθcosθ
= sin2次θtanθ-2sin 2次θ(cosθ/sinθ)+2sinθcosθ
= sin2次θtanθ-2sinθcosθ+2sinθcosθ
= sin2次θtanθ