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