中三 math 三角恆等恆問題

1. tanθsinθ over tanθ+ sinθ=tanθ- sinθ over tanθsinθ
tanθsinθ/(tanθ+sinθ)=(tanθ-sinθ)/(tanθsinθ)
sol
(tanθ-sinθ)(tanθ-sinθ)
=tan^2θ-sin^2θ
=sin^2θ/cos^2θ-sin^2θ
=sin^2θ(sec^2θ-1)
=sin^2θtan^2θ
=tanθsinθtanθsinθ
so tanθsinθ/(tanθ+sinθ)=(tanθ-sinθ)/(tanθsinθ)

2. cosθ- sinθ over 1 + 2sinθcosθ= 1 - tanθ over 1 + tanθ
(cos^2θ-sin^2θ)/(1+2sinθcosθ)=(1-tanθ)/(1+tanθ)
sol
(1-tanθ)/(1+tanθ)
=(cosθ-sinθ)/(cosθ+sinθ)
=(cosθ-sinθ)(cosθ+sinθ)/(cosθ+sinθ)^2
=(cos^2θ-sin^2θ)/(cos^2θ+2sinθcosθ+cos^2θ)
=(cos^2θ-sin^2θ)/(1+2sinθcosθ)

解答在http://hk.knowledge.yahoo.com/question/question?qid=7009032900618