已知tanθ+sinθ=a,tanθ-sinθ=b,求证:(a2-b2)2=16ab.

问题描述:

已知tanθ+sinθ=a,tanθ-sinθ=b,求证:(a2-b22=16ab.

证明:∵(a2-b22=[(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-b22=16ab.