## Trigonometry (11th Edition) Clone

$$\tan195^\circ=2-\sqrt 3$$
$$A=\tan195^\circ$$ We can separate $195^\circ$ into $180^\circ$ and $15^\circ$. Therefore, we can rewrite A as follows $$A=\tan(180^\circ+15^\circ)$$ $$A=\frac{\tan180^\circ+\tan15^\circ}{1-\tan180^\circ\tan15^\circ}$$ As $\tan180^\circ=\frac{\sin180^\circ}{\cos180^\circ}$, and $\sin180^\circ=0$ and $\cos180^\circ=-1$, $\tan180^\circ=0$. Which means $$A=\frac{0+\tan15^\circ}{1-0\times\tan15^\circ}$$ $$A=\frac{\tan15^\circ}{1}$$ $$A=\tan15^\circ$$ Since $15^\circ$ is half of $30^\circ$, we can rewrite $$A=\tan\Bigg(\frac{1}{2}30^\circ\Bigg)$$ $$A=\frac{\sin30^\circ}{1+\cos30^\circ}$$ $$A=\frac{\frac{1}{2}}{1+\frac{\sqrt 3}{2}}$$ $$A=\frac{\frac{1}{2}}{\frac{2+\sqrt 3}{2}}$$ $$A=\frac{2}{2(2+\sqrt 3)}$$ $$A=\frac{1}{2+\sqrt 3}$$ $$A=\frac{1}{2+\sqrt 3}\frac{2-\sqrt 3}{2-\sqrt 3}$$ $$A=\frac{2-\sqrt 3}{4-3}$$ $$A=2-\sqrt 3$$