1.已知sina,cosa为方程2x的平方+(根号3-1)x+m=0的两个不等根,其中0

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

因为sinA+cosA=(√3 -1)/2
sinAcosA=m/2
又因为sin平方A+cos平方A=1 =（sinA+cosA）^2-2sinAcosA
=(4-2√3)/4-2m=1 解得m=-√3/2
因为sinAcosA=m/2=-√3/4
又因为sinAcosA/1=-√3/4
将上式等号左边分母1换成sin平方a+cos平方...

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
...

有根与系数关系知:sinA*cosA=(1-√"3)/2，sinAXcosA=m/2，由前式的平方，后式代入求m,然后求