Answer
The identity is verified.
$cos~x-\frac{cos~x}{1-tan~x}=\frac{sin~x~cos~x}{sin~x-cos~x}$
Work Step by Step
Remember:
$tan~x=\frac{sin~x}{cos~x}$
So:
$cos~x~tan~x=cos~x~\frac{sin~x}{cos~x}=sin~x$
$cos~x-\frac{cos~x}{1-tan~x}=cos~x~\frac{1-tan~x}{1-tan~x}-\frac{cos~x}{1-tan~x}=\frac{cos~x-sin~x-cos~x}{1-tan~x}=\frac{sin~x}{tan~x-1}~\frac{cos~x}{cos~x}=\frac{sin~x~cos~x}{sin~x-cos~x}$
