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.

`=sqrt((1-cos a)^2/(1-cos^2a))`

We then make use of the identity `sin^2theta+cos^2theta=1`

`=sqrt((1-cos a)^2/(sin^2a))`

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.