Answer
$\cot(\frac{\pi}{2}-u)=\tan u$
Work Step by Step
Start with the left side:
$\cot(\frac{\pi}{2}-u)$
Express cotangent as cosine divided by sine:
$=\frac{\cos (\frac{\pi}{2}-u)}{\sin(\frac{\pi}{2}-u)}$
Expand using the subtraction formulas for sine and cosine:
$=\frac{\cos \frac{\pi}{2}\cos u+\sin \frac{\pi}{2}\sin u}{\sin \frac{\pi}{2}\cos u-\cos \frac{\pi}{2}\sin u}$
Evaluate $\sin \frac{\pi}{2}$ and $\cos \frac{\pi}{2}$:
$=\frac{0*\cos u+1*\sin u}{1*\cos u-0*\sin u}$
$=\frac{\sin u}{\cos u}$
$=\tan u$
Since this equals the right side, the identity has been proven.
