(1-cos^4X-sin^4x)∕(1-cos^6x-sin^6x)=化简(急)

化简 1-sin^4x-cos^4x/1-sin^6x-cos^6x 化简(1-cos^4x-sin^4x)/(1-cos^6x-sin^6x) 求值：(1-sin^6 x-cos^6 x)/(1-sin^4 x-cos^4 x) 化简[1-(sin^4x-sin^2cos^2x+cos^4x)/(sin^2)]+3sin^2x (1-cos^4X-sin^4x)∕(1-cos^6x-sin^6x)=化简(急) 求证（cos^2 x-sin^2 x)(cos^4 x+sin^4 x)+1/4 sin 2x sin 4x=cos 2x (1-(sin^4x-sin^2xcos^2x+cos^4x)/sin^2x +3sin^2x 化简[1-(sin^4 x-sin^2 xcos^2 x+cos^4 x)]/(sin^2 x)+3sin^2 x cos x/sin x+sin x/(1+cos x) 化简. 证明一个sin&cos的等式证明 1+sin^2(x)+sin^4(x)+sin^6(x)=[1-sin^8(x)]/[cos^2(x)] 求证(3-sin^4 x-cos^4 x)/2cos^2 x=1+tan^2 x+sin^2 x 求∫1/(sin^4x+cos^4x)dx, ∫[1/(sin^2(x)cos^4(x)]dx 求解∫1/(cos^4(x)sin^2(x))dx 求证:(sin 2x /(1-cos 2x) )·(sin x /(1+sin x))=tan (π/4-x/2). 求证（cos^2x-sin^2x)(cos^4x+sin^4x）+1/4sin2xsin4x=cos2x 求(1-sin^6x-cos^6x)/(1-sin^4x-cos^4x)的值 求1-cos^6x-sin^6x/1-cos^4x-sin^4x的值,