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-b2)2=[(a+b)(a-b)]2=[(tanθ+sinθ+tanθ-sinθ)(tanθ+sinθ-tanθ+sinθ)]2=16tan2θsin2θ.又16ab=16(tan2θ-sin2θ)=16•sin2θsin2θcos2θ=16•tan2θsin2θ.故有(a2-b2)2=16ab....