赞 nightwolflz 幼苗
共回答了24个问题采纳率：87.5% 举报
(1-sin^6 a-cos^6 a)/(1-sin^4 a-cos^4 a)
=[1-(sin^6 a+cos^6 a)]/[(1-sin^4 a-cos^4 a]
=[1-(sin^2 a+cos^2 a)(sin^4 a-sin^2 acos^2 a+cos^4 a)]/[(1-sin^4 a-cos^4 a]
=[1-(sin^4 a-sin^2 acos^2 a+cos^4 a)]/(1-sin^4 a-cos^4 a)
=[1-sin^4 a+sin^2 acos^2 a-cos^4 a]/(1-sin^4 a-cos^4 a)
=1+sin^2 acos^2 a/(1-sin^4 a-cos^4 a)
=1+sin^2 acos^2 a/[(1-sin^4 a)-cos^4 a]
=1+sin^2 acos^2 a/[(1-sin^2 a)(1+sin^2 a)-cos^4 a]
=1+sin^2 acos^2 a/[cos^2 a(1+sin^2 a)-cos^4 a]
=1+sin^2 acos^2 a/[cos^2 a(1+sin^2 a-cos^2a)]
=1+sin^2 acos^2 a/[2sin^2 a*cos^2 a]
=1+1/2
=3/2
=[1-(sin^6 a+cos^6 a)]/[(1-sin^4 a-cos^4 a]
=[1-(sin^2 a+cos^2 a)(sin^4 a-sin^2 acos^2 a+cos^4 a)]/[(1-sin^4 a-cos^4 a]
=[1-(sin^4 a-sin^2 acos^2 a+cos^4 a)]/(1-sin^4 a-cos^4 a)
=[1-sin^4 a+sin^2 acos^2 a-cos^4 a]/(1-sin^4 a-cos^4 a)
=1+sin^2 acos^2 a/(1-sin^4 a-cos^4 a)
=1+sin^2 acos^2 a/[(1-sin^4 a)-cos^4 a]
=1+sin^2 acos^2 a/[(1-sin^2 a)(1+sin^2 a)-cos^4 a]
=1+sin^2 acos^2 a/[cos^2 a(1+sin^2 a)-cos^4 a]
=1+sin^2 acos^2 a/[cos^2 a(1+sin^2 a-cos^2a)]
=1+sin^2 acos^2 a/[2sin^2 a*cos^2 a]
=1+1/2
=3/2
1年前
3
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