Answer
$\frac{1-cos~x}{sin~x} = tan~\frac{x}{2}$
Work Step by Step
$\frac{1-cos~x}{sin~x}$
When we graph this function, we can see that it looks like the graph of $~~tan~\frac{x}{2}$
We can verify this algebraically:
$\frac{1-cos~x}{sin~x} = \frac{1-cos~(\frac{x}{2}+\frac{x}{2})}{sin~(\frac{x}{2}+\frac{x}{2})}$
$\frac{1-cos~x}{sin~x} = \frac{1-(cos~\frac{x}{2}~cos~\frac{x}{2}-sin~\frac{x}{2}~sin~\frac{x}{2})}{sin~\frac{x}{2}~cos~\frac{x}{2}+cos~\frac{x}{2}~sin~\frac{x}{2}}$
$\frac{1-cos~x}{sin~x} = \frac{(1-cos^2~\frac{x}{2})+sin^2~\frac{x}{2}}{2~sin~\frac{x}{2}~cos~\frac{x}{2}}$
$\frac{1-cos~x}{sin~x} = \frac{sin^2~\frac{x}{2}+sin^2~\frac{x}{2}}{2~sin~\frac{x}{2}~cos~\frac{x}{2}}$
$\frac{1-cos~x}{sin~x} = \frac{2~sin^2~\frac{x}{2}}{2~sin~\frac{x}{2}~cos~\frac{x}{2}}$
$\frac{1-cos~x}{sin~x} = \frac{sin~\frac{x}{2}}{cos~\frac{x}{2}}$
$\frac{1-cos~x}{sin~x} = tan~\frac{x}{2}$
