Prove identities

應該是：tanθ/[1+(tanθ)^2]=sinθcosθ ？

LHS = tanθ/[1+(tanθ)^2]
= tanθ/[1+(sinθ/cosθ)^2]
= tanθ/ {[(cosθ)^2+(sinθ)^2]/(cosθ)^2}
= tanθ/ [1/(cosθ)^2]
= tanθ(cosθ)^2
=(sinθ/cosθ)(cosθ)^2
=sinθcosθ = RHS


2011-04-20 15:23:38 補充：
簡化版：

LHS = tanθ/[1+(tanθ)^2]
= tanθ/(secθ)^2
= tanθ(cosθ)^2
=(sinθ/cosθ)(cosθ)^2
=sinθcosθ = RHS