化简(sin^2x-cos^4x+cos^2x-sin^4x)/(sin^2x-cos^6x+cos^2x-sin^6x)
问题描述：
化简(sin^2x-cos^4x+cos^2x-sin^4x)/(sin^2x-cos^6x+cos^2x-sin^6x)
答
(sin^2x-cos^4x+cos^2x-sin^4x)/(sin^2x-cos^6x+cos^2x-sin^6x)=[sinx^2(1-sinx^2)+cosx^2(1-cosx^2)]/[sinx^2(1-sinx^4)+cosx^2(1-cosx^4)]=2sinx^2cosx^2/[3*sinx^2cosx^2]=2/3√希望你能看懂,你能明白, 望采纳,赞...