Evaluate $\tan 195^{\circ}$ without using the calculator How to evaluate $\tan 195^{\circ}$ without using the calculator, and how to give the answer in the form $a+b \sqrt{3}$, where $a$ and $b$ are integers?
 A: Hint: Please note that $$195=180+45-30.$$
A: Using the identity $\tan \left(180^\circ+\theta\right)=\tan\theta$ and $\tan \left(\alpha-\beta\right)=\dfrac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}$,
$\quad\tan195^\circ \\=\tan 15^\circ\\=\tan \left(60^\circ-45^\circ\right)\\=\dfrac{\tan60^\circ-\tan45^\circ}{1+\tan60^\circ\tan45^\circ}\\=\dfrac{\sqrt{3}-1}{\sqrt{3}+1}\\=\dfrac{\left(\sqrt{3}-1\right)\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\\=\dfrac{4-2\sqrt{3}}{2}\\=\boxed{2-\sqrt{3}}$
A: This may be the most inefficient way to do this, but a thorough method would be through the compound angle forumulae:


A: The Tangent Identities:
$$\tan (2\theta) = \frac{2\tan\theta}{1-\tan^2\theta} $$
$$\tan (\alpha + \beta) = \frac{\tan\alpha + \tan\beta}{1-\tan\alpha\tan\beta} $$
$$\tan (\alpha - \beta) = \frac{\tan\alpha - \tan\beta}{1+\tan\alpha\tan\beta} $$
$$\tan 195^\circ = \tan (180^\circ + 15^\circ)$$
$$\tan(180^\circ) = 0$$
$$\frac{0 + \tan(15^\circ)}{1 - 0 (\tan(15^\circ))} = \tan(15^\circ)$$
$$\tan(15^\circ) = \tan(45^\circ - 30^\circ)$$
$$\tan (45^\circ - 30^\circ) = \frac{\tan(45^\circ) - \tan(30^\circ)}{1+\tan(45^\circ)\tan(30^\circ)} $$
$$\tan(45^\circ) = 1$$
$$\tan(30^\circ) = \frac{1}{\sqrt3}$$
$$\frac{1 - \frac{1}{\sqrt3}}{1+\frac{1}{\sqrt3}}$$
A: As $\tan(180^\circ n+x)=\tan x, \tan(180^\circ +15^\circ)=\tan15^\circ$
As for $\sin y\ne0,$ 
$$\tan y=\dfrac{2\sin^2y}{2\sin y\cos y}=\dfrac{1-\cos2y}{\sin2y}$$
$$\tan15^\circ=\dfrac{1-\cos30^\circ}{\sin30^\circ}=?$$
A: Let's build a general formula, from $\tan(x)\text{ to }t = \tan({x\over2})$
$$\tan(x) = {2t \over 1-t^2}$$
$$t^2 + 2\cot(x) t -1 = 0$$
$$t = -\cot(x) ± \sqrt{\cot^2(x)+1}$$
Since $\tan(195°) = \tan(15°) > 0 \text{, and }\cot(30°)=\sqrt3$
$$\tan(195°) = \tan{30°\over2}= -\sqrt3 + \sqrt{3+1} = 2-\sqrt3$$

BTW, the "throw-away" solution is $\tan(15° ± 90°) = -2-\sqrt3$
