Answer
The trigonometric Identity is verified.
$\frac{cos(\frac{\pi}{2}-x)}{sin(\frac{\pi}{2}-x)}=tan~x$
Work Step by Step
We know that:
$cos(\frac{\pi}{2}-x)=sin~x$
$sin(\frac{\pi}{2}-x)=cos~x$ (Cofunction Identities, page 508)
Start at the left side of the equation:
$\frac{cos(\frac{\pi}{2}-x)}{sin(\frac{\pi}{2}-x)}=\frac{sin~x}{cos~x}=tan~x$
