# Tangent Half-Angle Identitiy

1. Nov 5, 2009

### Yuqing

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.

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

2. Nov 5, 2009

### tiny-tim

Hi Yuqing!

(have a theta: θ )

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

3. Nov 5, 2009

### Yuqing

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

Thank you very much.