Answer
Since $\tan(\alpha-\beta)=\dfrac{\tan(\alpha)-\tan(\beta)}{1+\tan(\alpha)\tan(\beta)}$ and $\tan{(2\pi)}=0$, then
$\tan(2\pi-\theta)=\dfrac{\tan(2\pi)-\tan(\theta)}{1+\tan(2\pi)\tan(\theta)}=\dfrac{0-\tan(\theta)}{1+0\tan(\theta)}=\dfrac{-\tan(\theta)}{1}=-\tan(\theta)$
Thus,
$\tan{(2\pi-\theta)}=-\tan{\theta}$
Work Step by Step
Recall:
$\tan(\alpha-\beta)=\dfrac{\tan(\alpha)-\tan(\beta)}{1+\tan(\alpha)\tan(\beta)}$
Hence,
$\tan(2\pi-\theta)\\
=\dfrac{\tan(2\pi)-\tan(\theta)}{1+\tan(2\pi)\tan(\theta)}\\
=\dfrac{0-\tan(\theta)}{1+0\tan(\theta)}\\
=\dfrac{-\tan(\theta)}{1}\\
=-\tan(\theta)$
