已知sinα+coaα=√2/2,1/sinα^2+1/cosα^2

1,sinα+coaα=√2/2
(sinα+coaα)^2=1/2
1+2sinαcosα=1/2
sinαcosα=-1/4
(sinαcosα)^2=1/16
1/sinα^2+1/cosα^2
=1/((sinαcosα)^2)=16
2,已知tanα=-1/2,求sinα^2+2sinαcosα-3cosα^2
sinα^2+2sinαcosα-3cosα^2
=(sinα^2+2sinαcosα-3cosα^2)/(sinα^2+cosα^2)
=(tanα^2+2tanα-3)/(tanα^2+1)
=(1/4-1-3)/(1/4+1)
=(-15/4)/(5/4)
=-3
3,求证：1-2sinαcoaα / sinα^2-cosα^2 = tanα+1 / tanα-1
∵cosa^2+sina^2=1,
∴左边=-[cosa^2+sina^2-2sina*cosa/cos^2-sin^2]
=-(cosa-sina)^2/cos^2-sin^2
上下同除cosa-sina
原式=-（cosa-sina）/cosa+sina
上下同除cosa
原式=（tana-1）/(1+tana)

1/sinα^2+1/cosα^2 =1/[(sina)^2(cosa)^2]即[1/sinacosa]^2
再将sinα+coaα=√2/2两边平方得
1+2sinacosa=1/2得sinacosa=-(1/4)
则[1/sinacosa]^2=16

16

sinα+coaα=√2/2
sin²α+2sinacosa+coa²α=1/2
1+2sinacosa=1/2
sinacosa=-1/4
1/sinα^2+1/cosα^2
=1/[sin²acos²a]
=16