Trig mind-bender!

$\tan \frac{13\pi}{12} = \tan (\pi+\frac{\pi}{12}) = \tan \frac{\pi}{12} = \tan (\frac{\pi}{3}-\frac{\pi}{4})$ $= \frac{\tan \frac{\pi}{3}-\tan \frac{\pi}{4}}{1+\tan\frac{\pi}{3}\tan \frac{\pi}{4}}$ $= \frac{\sqrt{3}-1}{1+\sqrt{3}.1} = \frac{\sqrt{3}-1}{\sqrt{3}-1}$ $= \frac{\sqrt{3}-1}{\sqrt{3}+1} . \frac{\sqrt{3}-1}{\sqrt{3}-1}$ $= \frac{(\sqrt{3}-1)^2}{3-1}$ $= \frac{3+1-2\sqrt{3}}{2}$ $= 2-\sqrt{3}$