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

find the exact value of $\mathrm{tan}\left(\frac{4\pi }{3}-\frac{\pi }{4}\right)$

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Alix Ortiz

Trigonometry:

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

We know

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

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

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

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

$=\sqrt{3}$

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

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

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

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

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

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

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

$=2-\sqrt{3}$

$=0.2679$