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