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Tangent Half-Angle Identitiy

  1. Nov 5, 2009 #1
    I saw this identity somewhere and have been looking for a derivation but I can't seem to find one. It would be of great help if someone can show me where this comes from.

    [tex]tan\frac{\theta}{2} = csc\theta - cot\theta [/tex]
  2. jcsd
  3. Nov 5, 2009 #2


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    Hi Yuqing! :smile:

    (have a theta: θ :wink:)

    Hint: cscθ - cotθ = (1 - cosθ)/sinθ = … ? :smile:
  4. Nov 5, 2009 #3
    Ah I see. Didn't think it was so simple.

    Thank you very much.
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