Answer
The identity is verified.
$cot~x-tan~x=sec~x(csc~x-2~sin~x)$
Work Step by Step
$cos^2x+sin^2x=1$
$cos^2x=1-sin^2x$
$cot~x-tan~x=\frac{cos~x}{sin~x}-\frac{sin~x}{cos~x}=\frac{cos~x}{sin~x}\frac{cos~x}{cos~x}-\frac{sin~x}{cos~x}\frac{sin~x}{sin~x}=\frac{cos^2x}{sin~x~cos~x}-\frac{sin^2x}{cos~x~sin~x}=\frac{cos^2x-sin^2x}{cos~x~sin~x}=\frac{1-sin^2x-sin^2x}{cos~x~sin~x}=\frac{1-2~sin^2x}{cos~x~sin~x}=\frac{1}{cos~x~sin~x}-\frac{2~sin^2x}{cos~x~sin~x}=sec~x~csc~x-\frac{1}{cos~x}~(2~sin~x)=sec~x~csc~x-sec~x~(2~sin~x)=sec~x(csc~x-2~sin~x)$
