Question

已知sinx，cosx是关于x的二次方程2x^2+(√2+1)x+m=0的两根。求cosx/（1-cotx^2）+sinx/（1-tanx^2）的值

请写出详细步骤
关键是后面那个式子不知道怎么化

Answer 1

2x^2+(√3+1)x+m=0
sinx+cosx= -(√3+1)/2
sinxcosx=m/2
(sinx)^2+(cosx)^2
=(sinx+cosx)^2-2sinxcosx
=(4+2√3)/4-m=1
m=√3/2
2x^2+(√3+1)x+√3/2=0
(2x+1)(x+√3/2)=0
x=-1/2,或x=-√3/2
sinx=-1/2,cosx=-√3/2 或 sinx=-√3/2,cosx=-1/2
cosx/(1-(cotx)^2)=cosx/[1-(cosx/sinx)^2] =(sinx)^2cosx/[(sinx)^2-(cosx)^2]
sinx/(1-(tanx)^2)=(cosx)^2sinx/[(cosx)^2-(sinx)^2]
cosx/(1-(cotx)^2)+sinx/(1-(tanx)^2)=(cosx-sinx)sinxcosx/[(cosx)^2-(sinx)^2]
=sinxcosx/(sinx+cosx)
=(-1/2)(-√3/2)/(-1/2-√3/2)=2√3/(1+√3)=√3(√3-1)=3-√3

8x^2+(4√2+4)x+(2√2-1)=0
(2x+1)(4x+2√2-1)=0
sinx=-1/2,cosx= -(2√2-1)/4

Answer 2

cosx/（1-cotx^2）+sinx/（1-tanx^2）
=cosxtanx^2/(tanx^2-1)+sinx/(1-tanx^2)
=(sinx-cosxtanx^2)/(1-tanx^2)
=sinx(1-tanx)/(1+tanx)(1-tanx)
=sinx/(1+tanx)=sinxcosx/(cosx+sinx)
后面会做了吧