Question

[(a+2/a^-2a)-(a-1/a^2-4a+4)]*(a-2/a)/(4-a/a^2)

Answer 1

(cosα)
=f(sin(π/2-α))
=2-cos[2(π/2-α)]
=2-cos(π-2α)
=2+cos2α
判别式大于0
4k+4>0
k>-1

k在根号下
所以k≥0

Answer 2

原式=[(a+2)/a(a-2)-(a-1)/(a-2)²]*[(a-2)/a*a²/(4-a)]
=[(a²-4-a²+a)/a(a-2)²]*[-a(a-2)/(a-4)]
=[(a-4)/a(a-2)²]*[-a(a-2)/(a-4)]
=-a(a-2)(a-4)/[a(a-4)(a-2)²]
=-1/(a-2)