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

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

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Alix Ortiz

Trigonometry:

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

We know

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

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

$$=\tan(90\cdot2+60^\circ)$$

$$=\tan60^\circ$$

$$=\sqrt{3}$$

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

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

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

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

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

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

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

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

$$=0.2679$$