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