Answer
$sec~x~tan~x-sin~x=tan^2x~sin~x$
Work Step by Step
$sec~x~tan~x-sin~x=tan~x(sec~x-\frac{sin~x}{tan~x})=tan~x(\frac{1}{cos~x}-sin~x~\frac{cos~x}{sin~x})=tan~x(\frac{1}{cos~x}-cos~x)=tan~x(\frac{1-cos^2x}{cos~x})=tan~x~\frac{sin^2x}{cos~x}=tan~x~\frac{sin~x}{cos~x}~sin~x=tan^2x~sin~x$
