## Algebra and Trigonometry 10th Edition

$tan^2x-tan^2x~sin^2x=sin^2x$
Pythagorean Identity: $cos^2x+sin^2x=1$ $cos^2x=1-sin^2x$ $tan^2x-tan^2x~sin^2x=tan^2x(1-sin^2x)=\frac{sin^2x}{cos^2x}cos^2x=sin^2x$