求证sin2α/(sinα-cosα)-(sinα+cosα)/(tan2-1)=sinα+cosα(1)sin2α/(sinα-cosα)-(sinα+cosα)/(tan2-1)=sinα+cosα(2)(2-cos2α)×(2+tan2α)=(1+2tan2α)×(2-sin2α)

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求证sin2α/(sinα-cosα)-(sinα+cosα)/(tan2-1)=sinα+cosα
(1)sin2α/(sinα-cosα)-(sinα+cosα)/(tan2-1)=sinα+cosα(2)(2-cos2α)×(2+tan2α)=(1+2tan2α)×(2-sin2α)

(1)sin2α/(sinα-cosα)-(sinα+cosα)/(tan2-1)=sin2α/(sinα-cosα)-cos2α/(sinα-cosα)=(sin2α-cos2α)/(sinα-cosα)=sinα+cosα(2)左边-右边=(1+sin2α)(2+tan2α)-(1+2tan2α)(2-sin2α)=(2+2sin2α+tan2α+sin2αtan2α)-(2-sin2α+4tan2α-2sin2αtan2α)=3sin2α-3tan2α+3sin2αtan2α=3sin2α-3tan2α(1-sin2α)=3sin2α-3sin2α=0