## Algebra and Trigonometry 10th Edition

False. $tan~x=cot(\frac{\pi}{2}-x)$ It is also required to use a Pythagorean Identity.
$tan~x=cot(\frac{\pi}{2}-x)$ $tan^2x=cot^2(\frac{\pi}{2}-x)$ Use the Pythagorean Identity: $csc^2u=cot^2u+1$ $cot^2u=csc^2u-1$ $tan^2x=csc^2(\frac{\pi}{2}-x)-1$ $tan~x=\pm\sqrt {csc^2(\frac{\pi}{2}-x)-1}$