Tangent Half-Angle Identitiy

**Yuqing** · Nov5-09, 06:18 PM

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.

$LaTeX Code: tan\\frac{\\theta}{2} = csc\\theta - cot\\theta$

**tiny-tim** · Nov5-09, 06:54 PM

Hi Yuqing!

(have a theta: θ

)

Hint: cscθ - cotθ = (1 - cosθ)/sinθ = … ?

**Yuqing** · Nov5-09, 07:39 PM

Ah I see. Didn't think it was so simple.

Thank you very much.