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