解tan2θ-tanθ-tanθ
=tan2θ-2tanθ
=2tanθ/(1-tan^2θ)-2tanθ
=2tanθ[1/(1-tan^2θ)-1]
=2tanθ[（1-(1-tan^2θ))/(1-tan^2θ)]
=2tanθ[tan^2θ/(1-tan^2θ)]
=2tan^3θ/(1-tan^2θ)
知tanθ＞1时,2tan^3θ/(1-tan^2θ)＜0,即tan2θ-tanθ＜tanθ,
当0＜tanθ＜1时,2tan^3θ/(1-tan^2θ)＞0,即tan2θ-tanθ＞tanθ
当-1＜tanθ＜0时,2tan^3θ/(1-tan^2θ)＜0,即tan2θ-tanθ＜tanθ
当tanθ＜-1时,2tan^3θ/(1-tan^2θ)＞0,即tan2θ-tanθ＞tanθ