(1-cos4α)/sin4α*cos2α/(1+cos2α)
=2(sin2α)^2/(2sin2αcos2α)*cos2α/(1+cos2α)[(sin2α)^2表示sin2α的平方]
=sin2α/(1+cos2α)
=2sinαcosα/[2(cosα)^2]
=tanα