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