答:
x=2-√3代入方程得:
(2-√3)^2-(tanθ+cotθ)*(2-√3)+1=0
(2-√3)(tanθ+cotθ)=7-4√3+1
tanθ+cotθ=4(2-√3)/(2-√3)=4
所以:
sinθ/cosθ+cosθ/sinθ=4
[ (sinθ)^2+(cosθ)^2 ]/ (sinθcosθ)=4
1/ (sinθcosθ)=4
sinθcosθ=1/4
答:
x=2-√3代入方程得:
(2-√3)^2-(tanθ+cotθ)*(2-√3)+1=0
(2-√3)(tanθ+cotθ)=7-4√3+1
tanθ+cotθ=4(2-√3)/(2-√3)=4
所以:
sinθ/cosθ+cosθ/sinθ=4
[ (sinθ)^2+(cosθ)^2 ]/ (sinθcosθ)=4
1/ (sinθcosθ)=4
sinθcosθ=1/4