解:tanθ+cotθ=(sinθ/cosθ)+(cosθ/sinθ)=[(sinθ)^2+(cosθ)^2]/(sinθcosθ)=2/sin2θ(sinθ+cosθ)^2=(sinθ)^2+(cosθ)^2+2sinθcosθ=1+sin2θ=(√2)^2=2故sin2θ=1代入有:tanθ+cotθ=2/sin2θ=2/1=2
