Answer
$sec~\theta-sin~\theta~tan~\theta= cos~\theta$
Work Step by Step
We can express $sec~\theta-sin~\theta~tan~\theta$ as a single function of $\theta$:
$sec~\theta-sin~\theta~tan~\theta$
$= \frac{1}{cos~\theta}-sin~\theta~\frac{sin~\theta}{cos~\theta}$
$= \frac{1}{cos~\theta}-\frac{sin^2~\theta}{cos~\theta}$
$= \frac{1-sin^2~\theta}{cos~\theta}$
$= \frac{cos^2~\theta}{cos~\theta}$
$= cos~\theta$
