Question

已知tanθ=2则(sinθ+2cosθ)/(sinθ-cosθ)的值为

Answer 1

(sinθ+2cosθ)/(sinθ-cosθ) (上下同除以cosθ)
=(tanθ+2)/(tanθ-1)
=4

追问

那(sin^2θ+2cosθsinθ+1)=？

追答

(sin^2θ+2cosθsinθ+1)

=(sin^2θ+2cosθsinθ)/1+1
=(sin^2θ+2cosθsinθ)/(sin^2θ+cos^2θ) +1 (上下同除以cos^2θ)
=(tan^2θ+2tanθ)/(tan^2θ+1) +1
=8/5+1
=13/5

Answer 2

解
原式
=(tanθ+2)/(tanθ-1)——分子分母同时除以cosθ
=(2+2)/(2-1)
=4

追问

那(sin^2θ+2cosθsinθ+1)=？

追答

解
(sin²θ+2cosθsinθ+1)
=(sin²θ+2cosθsinθ+sin²θ+cos²θ)/(sin²θ+cos²θ)——除以sin²θ+cos²θ=1，值不变
=(2sin²θ+2cosθsinθ+cos²θ)/(sin²θ+cos²θ)——分子分母同时除以cos²θ
=(2tan²θ+2tanθ+1)/(tan²θ+1)
=(8+4+1)/(4+1)
=13/5

Answer 3

原式=(tanθ+2)/(tanθ-1)
=4

追问

那(sin^2θ+2cosθsinθ+1)=？

追答

13/5