## Algebra and Trigonometry 10th Edition

$cos~x(sec~x-cos~x)=sin^2x$
We know that: $sec~x=\frac{1}{cos~x}$ $sin^2x+cos^2x=1$ $sin^2x=1-cos^2x$ So: $cos~x(sec~x-cos~x)=cos~x(\frac{1}{cos~x}-cos~x)=1-cos^2x=sin^2x$