∵tanθ=2，∴原式= 1?cos(2θ? π 2 )+cos2θ 1+ 1+cos2θ 2 = 2?2sin2θ+2cos2θ 3+cos2θ = 4cos2θ?4sinθcosθ 4cos2θ+2sin2θ = 4?4tanθ 4+2tan2θ = 4?8 4+8 =- 1 3 ．故选：D．
