高一三角函数 ,已知(1+tanx)/(1-tanx)=3+2√2,求(sin^2x+√2sinxcosx-cos^2x

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  • (sin^2x+√2sinxcosx-cos^2x)/(sin^2x+2cos^2x)=(tan²x+√2tanx-1)/(tan²x+2)

    (1+tanx)/(1-tanx)=3+2√2

    tanx=√2/2

    (sin^2x+√2sinxcosx-cos^2x)/(sin^2x+2cos^2x)=(tan²x+√2tanx-1)/(tan²x+2)=1/5

    sinθ/(1-1/tanθ)+cosθ/(1-tanθ)=sinθ/(1-cosθ/sinθ)+cosθ/(1-sinθ/cosθ)=sinθ+cosθ=(√3+1)/2