find the exact value of tan(14π/3) using the unit circle

Trigonometry:

$\tan (\frac{4\pi }{3}-\frac{\pi }{4})$

We know

$\tan (a-b)=\frac{\tan a-\tan b}{1+\tan a\tan b}$

$\tan (\frac{4\pi }{3})=\tan ({240}^{\circ })$

$=\tan (90\cdot 2+{60}^{\circ })$

$=\tan {60}^{\circ }$

$=\sqrt{3}$

$\tan (\frac{\pi }{4})=1$

$\tan (\frac{4\pi }{3}-\frac{\pi }{4})$

$=\frac{\tan \frac{4x}{3}-\tan \frac{x}{4}}{1+\tan \frac{4x}{3}\tan \frac{x}{4}}$

$=\frac{\sqrt{3}-1}{1+\sqrt{3}}$

$=\frac{(\sqrt{3}-1)(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)}$

$=\frac{3-2\sqrt{3}+1}{3-1}$

$=\frac{4-2\sqrt{3}}{2}$

$=2-\sqrt{3}$

$=0.2679$