已知(sinγ)^2=(sinα)^2-sinαcosαtan(α-β),求证:(tanγ)^2=tanαtanβ
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已知(sinγ)^2=(sinα)^2-sinαcosαtan(α-β),求证:(tanγ)^2=tanαtanβ
![已知(sinγ)^2=(sinα)^2-sinαcosαtan(α-β),求证:(tanγ)^2=tanαtanβ](/uploads/image/z/17408284-52-4.jpg?t=%E5%B7%B2%E7%9F%A5%28sin%CE%B3%29%5E2%3D%28sin%CE%B1%29%5E2-sin%CE%B1cos%CE%B1tan%28%CE%B1-%CE%B2%29%2C%E6%B1%82%E8%AF%81%3A%28tan%CE%B3%29%5E2%3Dtan%CE%B1tan%CE%B2)
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.
=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.
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