how to show x = 1 / (1+2costheta)?

2021/01/13: 
昨天看到你在回答區問的, 我就補充了.
(1/2)tan(θ)(1+tan^2(θ/2))/(tan(θ)+tan(θ/2))
      = (1/2)tan(θ)sec^2(θ/2)/(tan(θ)+tan(θ/2)) 
      = (1/2)(1/cos^2(θ/2))/(1+tan(θ/2)/tan(θ)) 
      = (1/2)/(cos^2(θ/2)+cos^2(θ/2)tan(θ/2)/tan(θ))
請再看看還有哪裡寫得不夠清楚.
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作 線段OC⊥AB邊 於 C.
則 ∠COX = θ/2,
∴ CX/OX = sin(θ/2)
   OC/OX = cos(θ/2)
   (1/2)/OA = sin(3θ/2)
   OC/OA = cos(3θ/2)

∴ OC = cot(3θ/2)/2
   OX = cot(3θ/2)/(2cos(θ/2))
   CX = cot(3θ/2)tan(θ/2)/2
   AX = 1/2 - cot(3θ/2)tan(θ/2)/2
      = (1/2)(1-tan(θ/2)/tan(3θ/2))
      = (1/2)[1-tan(θ/2)(1-tan(θ)tan(θ/2))/(tan(θ)+tan(θ/2))]
      = (1/2)tan(θ)(1+tan^2(θ/2))/(tan(θ)+tan(θ/2))
      = (1/2)tan(θ)sec^2(θ/2)/(tan(θ)+tan(θ/2))
      = (1/2)(1/cos^2(θ/2))/(1+tan(θ/2)/tan(θ))
      = (1/2)/(cos^2(θ/2)+cos^2(θ/2)tan(θ/2)/tan(θ))
      = (1/2)/(cos^2(θ/2)+sin(θ/2)cos(θ/2)/tan(θ))
      = (1/2)/[(1+cos(θ))/2 +(1/2)sin(θ)/tan(θ)]
      = 1/(1+2cos(θ))

Hi, Mr. 老怪物
how do you get    = (1/2)/(cos^2(θ/2)+cos^2(θ/2)tan(θ/2)/tan(θ))?