Answer
$\frac{1}{2}~cot~\frac{x}{2}-\frac{1}{2}~tan~\frac{x}{2} = cot~x$
Work Step by Step
We can verify that the equation is an identity:
$\frac{1}{2}~cot~\frac{x}{2}-\frac{1}{2}~tan~\frac{x}{2}$
$=\frac{1}{2}\cdot \frac{cos~\frac{x}{2}}{sin~\frac{x}{2}}-\frac{1}{2}\cdot \frac{sin~\frac{x}{2}}{cos~\frac{x}{2}}$
$=\frac{1}{2}~\Big(\frac{cos~\frac{x}{2}}{sin~\frac{x}{2}}-\frac{sin~\frac{x}{2}}{cos~\frac{x}{2}}\Big)$
$=\frac{1}{2}~\Big(\frac{cos^2~\frac{x}{2}}{sin~\frac{x}{2}~cos~\frac{x}{2}}-\frac{sin^2~\frac{x}{2}}{sin~\frac{x}{2}~cos~\frac{x}{2}}\Big)$
$=\frac{1}{2}~\Big(\frac{cos^2~\frac{x}{2}-sin^2~\frac{x}{2}}{sin~\frac{x}{2}~cos~\frac{x}{2}}\Big)$
$=\frac{cos^2~\frac{x}{2}-sin^2~\frac{x}{2}}{2~sin~\frac{x}{2}~cos~\frac{x}{2}}$
$= \frac{cos~x}{sin~x}$
$= cot~x$
