Question

已知0<θ<π，且sinθ,cosθ是方程5x^2-x-12/5=0的两根，求sin^2θ-cos^2θ和tanθ+cotθ

Answer 1

sinθ,cosθ是方程5x^2-x-12/5=0的两根,根据根与系数的关系得：
sinθ+cosθ=1/5
sinθcosθ=-12/25 可得：cosθ<0
即：sinθ-cosθ=√[(sinθ+cosθ)^2-4sinθcosθ]
=7/5
sin^2θ-cos^2θ
=(sinθ+cosθ)(sinθ-cosθ)
=1/5x7/5
=7/25

tanθ+cotθ
=(sin^2θ+cos^2θ)/sinθcosθ
=1/sinθcosθ
=-25/12

Answer 2

25X^2-5X-12=0 (5X+3)(5X-4)=0 X1=4/5,X2=-3/5 因为a属于（0，π），sina>=0
所以，sina=4/5,coa=-3/5
(1)
sin^2a-cos^2a=7/25
(2)tana+cota=sina/cosa+cosa/sina=(sin^2a+cos^2a)/(sinacosa)=1/(-12/25)=-25/12