问题标题：
高一数学证明(sin2α/1+cotα)+(cos2α/1+tanα)=1-sinαcosα证明(sin2α/1+cotα)+(cos2α/1+tanα)=1-sinαcosα2是指数,二次方
问题描述：
高一数学证明(sin2α/1+cotα)+(cos2α/1+tanα)=1-sinαcosα
证明(sin2α/1+cotα)+(cos2α/1+tanα)=1-sinαcosα
2是指数,二次方
陈松林回答：
1
sin^2α/（1+cotα)+cos^2α/（1+tanα)
=sin^2α·sinα/（sinα+cosα)+cos^2α·cosα/（cosα+sinα)
=(sin^3α+cos^3α)/（sinα+cosα)
=(sinα+cosα)(sin^2α-sinα·cosα+cos^2α)/（sinα+cosα)
=sin^2α-sinα·cosα+cos^2α
=1-sinαcosα
2
tanαsina/(tanα-sina)
=tanαsina/(sinα/cosα-sina)
=(sinα/cosα)/(1/cosα-1)
=sinα/(1-cosα)
=1/tan(α/2);
(tanα+sinα)/tanαsinα
=((sinα/cosα)+sinα)/tanαsinα
=((1/cosα)+1)/tanα
=((1/cosα)+1)·cosα/tanα·cosα
=(1+cosα)/sinα
=1/tan(α/2);
∴左=右;
tanαsina/(tanα-sina)=(tanα+sinα)/tanαsinα
3
(1-sin^4α-cos^4α)/(1-sina^6α-cos^6α)
=[1-(sin^2α+cos^2α)^2+2sin^2α·cos^2α]/[1-(sin^2α+cos^2α)(sin^4α-sin^2α·cos^2α+cos^4α)]
=[1-1+2sin^2α·cos^2α]/[1-1*(sin^4α+2sin^2α·cos^2α+cos^4α-3sin^2α·cos^2α)]
=2sin^2α·cos^2α/[1-(sin^2α+cos^2α)^2+3sin^2α·cos^2α]
=2sin^2α·cos^2α/[1-1+3sin^2α·cos^2α]
=2/3
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