First, we recall `tan\ x = (sin x) / (cos x)`.
`tan\ a/2=(sin a/2)/(cos a/2)`
Then we use the sine and cosine of a half angle, as given above:
`=sqrt((1-cos a)/2)/sqrt((1+cos a)/2)`
Next line is the result of multiplying top and bottom by `sqrt 2`.
`=sqrt((1-cos a)/(1+cos a))`
We then multiply top and bottom (under the square root) by `(1 − cos α)`
`=sqrt(((1-cos a)^2)/((1+cos a)(1-cos a)))`
Next is a difference of 2 squares.
We then make use of the identity `sin^2theta+cos^2theta=1`
We then find the square root:
`=(1-cos a)/(sin a)`
Of course, we would need to make allowance for positive and negative signs, depending on the quadrant in question.