## Algebra and Trigonometry 10th Edition

$cos~x(1+tan^2x)=sec~x$
One of the Pythagorean Identities is: $1+tan^2x=sec^2x$ Also: $sec~x=\frac{1}{cos~x}~→~cos~x=\frac{1}{sec~x}$ $cos~x(1+tan^2x)=\frac{1}{sec~x}sec^2x=sec~x$