急问两个高一的数学问题!

1,
2x^2 + (3^(1/2) - 1)x + m = 0,

cosasina = m/2, m = sin(2a).

cosa + sina = (1 - 3^(1/2))/2,

sina = 1/2, cosa = -3^(1/2)/2, a = 5PI/6,

m = sin(5PI/3) = -3^(1/2)/2

tana = -3^(-1/2)

cota = -3^(1/2)

tana - cota = 3^(1/2) - 3^(-1/2) = 2/3^(1/2)

2,
sina = 2cosa,

(sina)^2 = 4(cosa)^2 = 4 - 4(sina)^2,

(sina)^2 = 4/5

3(sina)^2 - 2sinacosa = 3(sina)^2 - (sina)^2 = 2(sina)^2
= 2*4/5 = 8/5

sina,cosa为方程2x的平方+(根号3-1)x+m=0的两个不等根，则可知
2x的平方+(根号3-1)x+m==2(x-sina)(x-cosa)=0
因而，sina+cosa=-(根号3-1)/2 m=2sinacosa
两边平方得
(sina+cosa)^2
=(sin^2)a+(cos^2)a+2sinacosa
=1+m
=1-(√3)/2
得到m=-(√3)/2
tana-cota
=sina/cosa-cosa/sina
=((sina)^2-(cosa)^2)/(sinacosa)
=(sina+cosa)√((sina-cosa)^2)/(-√3/4)
=(-(√3-1)/2 )√(1-2sinacosa)/(-√3/4)
=±(2√3)/3

因为sinA+cosA=(√3 -1)/2
sinAcosA=m/2
又因为sin平方A