已知tanθ+sinθ=a,tanθ-sinθ=b,求证:(a2-b2)2=16ab.
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已知tanθ+sinθ=a,tanθ-sinθ=b,求证:(a2-b2)2=16ab.
![已知tanθ+sinθ=a,tanθ-sinθ=b,求证:(a2-b2)2=16ab.](/uploads/image/z/8477808-24-8.jpg?t=%E5%B7%B2%E7%9F%A5tan%CE%B8%2Bsin%CE%B8%3Da%EF%BC%8Ctan%CE%B8-sin%CE%B8%3Db%EF%BC%8C%E6%B1%82%E8%AF%81%EF%BC%9A%EF%BC%88a2-b2%EF%BC%892%3D16ab%EF%BC%8E)
证明:∵(a2-b2)2=[(a+b)(a-b)]2
=[(tanθ+sinθ+tanθ-sinθ)(tanθ+sinθ-tanθ+sinθ)]2
=16tan2θsin2θ.
又16ab=16(tan2θ-sin2θ)=16•
sin2θsin2θ
cos2θ=16•tan2θsin2θ.
故有(a2-b2)2=16ab.
=[(tanθ+sinθ+tanθ-sinθ)(tanθ+sinθ-tanθ+sinθ)]2
=16tan2θsin2θ.
又16ab=16(tan2θ-sin2θ)=16•
sin2θsin2θ
cos2θ=16•tan2θsin2θ.
故有(a2-b2)2=16ab.
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