求证!快 sin^2α/(1+cotα)+cos^2α/(1+tanα)=1-sinαcosα

2个回答

  • 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

    看看怎么样?