Question

已知tanθ=2,则sinθ/(sin3θ-cos3θ)=

A.10/7
B.7/10
C.9/7
D9/10

还有一种不用立方差的方法：
∵ tanθ=2
∴ sinθ=2cosθ, 代入sin²θ+cos²θ=15cos²θ=1
∴ cos²θ=1/5
∴ sinθ/(sin³θ-cos³θ)
=2cosθ/（8cos³θ-cos³θ）
=2cosθ/7cos³θ
=2/cos²θ
将cos²θ=1/5 即可得=10/7

Answer 1

解答：
∵ tanθ=2
∴ sinθ=2cosθ,
代入sin²θ+cos²θ=1
5cos²θ=1
∴ cos²θ=1/5
∴ sinθ/(sin³θ-cos³θ)
=sinθ/[(sinθ-cosθ)(sin²θ+sinθcosθ+cos²θ)]
=sinθ/[(sinθ-cosθ)(1+sinθcosθ)]
=2cosθ/[(2cosθ-cosθ)(1+2cosθcosθ)]
=2cosθ/[cosθ(1+2cos²θ)]
=2/(1+2cos²θ)
=2/(1+2/5)
=2/(7/5)
=10/7

Answer 2

tanθ=sinθ/cosθ=2
sinθ=2cosθ

sinθ/(sin^3θ-cos^3θ)
=sinθ/[(sinθ-cosθ)(sin^2θ+sinθcosθ+cos^2θ)]
=sinθ/[(sinθ-cosθ)(1+sinθcosθ)]
=sinθ/[(sinθ-cosθ)(1+sinθcosθ)]
=2cosθ/[(2cosθ-cosθ)(1+2cosθcosθ)]
=2cosθ/[cosθ(1+2cosθcosθ)]
=2/(2+2cos2θ)
=1/(1+cos2θ)
=1/[1+(1-tan^2θ)/(1+tan^2θ)]
=1/[1+(1-2^2)/(1+2^2)]
=1/[1-3/5]
=1/[2/5]
=5/2不懂追问，希望采纳
我这个答案肯定真确的
相信我