已知(sinγ)^2=(sinα)^2-sinαcosαtan(α-β),求证:(tanγ)^2=tanαtanβ

证明：∵sin²γ=sin²α-sinαcosαtan(α-β)
=sin²α-sinαcosαsin(α-β)/cos(α-β)
=sin²α-sinαcosα(sinαcosβ-cosαsinβ)/(cosαcosβ+sinαsinβ)
∴tan²γ=sin²γ/cos²γ
=sin²γ/(1-sin²γ)
=[sin²α-sinαcosα(sinαcosβ-cosαsinβ)/(cosαcosβ+sinαsinβ)]
/[1-sin²α+sinαcosα(sinαcosβ-cosαsinβ)/(cosαcosβ+sinαsinβ)]
=[sin²α(cosαcosβ+sinαsinβ)-sinαcosα(sinαcosβ-cosαsinβ)]
/[cos²α(cosαcosβ+sinαsinβ)+sinαcosα(sinαcosβ-cosαsinβ)]
=[sinαsinβ(sin²α+cos²α)]/[cosαcosβ(sin²α+cos²α)]
=(sinαsinβ)/(cosαcosβ)
=tanαtanβ.