若sin^x-sinxcosx-6cos^x=0,则[(cos2x-sin2x)/(1-cos2x)(1-tan2x)]=?
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若sin^x-sinxcosx-6cos^x=0,则[(cos2x-sin2x)/(1-cos2x)(1-tan2x)]=?
若sin^x-sinxcosx-6cos^x=0,则[(cos2x-sin2x)/(1-cos2x)(1-tan2x)]=?
若sin^x-sinxcosx-6cos^x=0,则[(cos2x-sin2x)/(1-cos2x)(1-tan2x)]=?
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