## Algebra and Trigonometry 10th Edition

The identity is verified. $sec^2x~cot~x-cot~x=tan~x$
$1+tan^2x=sec^2x~~$ (Pythagorean identity): $tan^2x=sec^2x-1$ $cot~x=\frac{1}{tan~x}$ $sec^2x~cot~x-cot~x=(sec^2x-1)cot~x=tan^2x~\frac{1}{tan~x}=tan~x$