[1-（sinα）^4-(cosα)^4]/[1-(sinα)^6-(cos)^6)]=?

[1-（sinα）^4-(cosα)^4]/[1-(sinα)^6-(cos)^6)]=?
[1-（sinα）^4-(cosα)^4]/[1-(sinα)^6-(cos)^6)]=?

[1-（sinα）^4-(cosα)^4]/[1-(sinα)^6-(cos)^6)]=?
因为sina^2+cosa^2=1
原式=[sin^2+cos^2-sin^4-cos^4]/[sin^2+cos^2-sina^6-cosa^6]
=[sin^2(1-sin^2)+cos^2(1-cos^2]/[sin^2(1-sin^4)+cos^2(1-cos^4]
=2*(sin^2)*(cos^2)/[sin^2*cos^2(1+sin^2+1+cos^2]
=2/3

因为sina^2+cosa^2=1
原式=[sin^2+cos^2-sin^4-cos^4]/[sin^2+cos^2-sina^6-cosa^6]
=[sin^2(1-sin^2)+cos^2(1-cos^2]/[sin^2(1-sin^4)+cos^2(1-cos^4]
=2*(sin^2)*(cos^2)/[sin^2*cos^2(1+sin^2+1+cos^2]
=2/3

=【(1-sin²α)(1+sin²α)-cos⁴α]/[(1-sin²α)(1+sin²α+sin⁴α)-(cosα)^6]
=[cos²α(1+sin²α)-cos⁴α]/[cos²α(1+sin²α+sin⁴α)-(cosα)^6]
=[cos
...

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=【(1-sin²α)(1+sin²α)-cos⁴α]/[(1-sin²α)(1+sin²α+sin⁴α)-(cosα)^6]
=[cos²α(1+sin²α)-cos⁴α]/[cos²α(1+sin²α+sin⁴α)-(cosα)^6]
=[cos²α(1+sin²α-cos²α)]/[cos²α(1+sin²α+sin⁴α-cos⁴α)]
=2sin²α/(1+sin²α+sin²α-cos²α)=2sin²α/3sin²α=2/3

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