Answer
$tan(\pi+x) = tan~x$
Work Step by Step
$tan(\pi+x) = \frac{tan~\pi+tan~x}{1-tan~\pi~tan~x}$
$tan(\pi+x) = \frac{\frac{sin~\pi}{cos~\pi}+\frac{sin~x}{cos~x}}{1-(\frac{sin~\pi}{cos~\pi})~(\frac{sin~x}{cos~x})}$
$tan(\pi+x) = \frac{\frac{0}{cos~\pi}+\frac{sin~x}{cos~x}}{1-(\frac{0}{cos~\pi})~(\frac{sin~x}{cos~x})}$
$tan(\pi+x) = \frac{\frac{sin~x}{cos~x}}{1}$
$tan(\pi+x) = tan~x$
