(tanθ-1)/(2tanθ+1)=1,则cos2θ/(1+sin2θ)=?
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![(tanθ-1)/(2tanθ+1)=1,则cos2θ/(1+sin2θ)=?](/uploads/image/z/2608757-53-7.jpg?t=%28tan%CE%B8-1%29%2F%282tan%CE%B8%2B1%29%3D1%2C%E5%88%99cos2%CE%B8%2F%281%2Bsin2%CE%B8%29%3D%3F)
(tanθ-1)/(2tanθ+1)=1,则cos2θ/(1+sin2θ)=?
(tanθ-1)/(2tanθ+1)=1,则cos2θ/(1+sin2θ)=?
(tanθ-1)/(2tanθ+1)=1,则cos2θ/(1+sin2θ)=?
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tan?=1/2 tan?=1/3 tan?=1/4 tan?=1/5 tan?=1/6