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