Prove that $\tan(75^\circ) = 2 + \sqrt{3}$ My (very simple) question to a friend was how do I prove the following using basic trig principles:
$\tan75^\circ = 2 + \sqrt{3}$
He gave this proof (via a text message!)
$1. \tan75^\circ$ 
$2. = \tan(60^\circ + (30/2)^\circ)$
$3. = (\tan60^\circ + \tan(30/2)^\circ) / (1 - \tan60^\circ \tan(30/2)^\circ) $
$4. \tan (30/2)^\circ = \dfrac{(1 - \cos30^\circ)}{ \sin30^\circ}$  
Can this be explained more succinctly as I'm new to trigonometry and a little lost after (2.) ?

EDIT
Using the answers given I'm almost there:


*

*$\tan75^\circ$   

*$\tan(45^\circ + 30^\circ)$  

*$\sin(45^\circ + 30^\circ) / \cos(45^\circ + 30^\circ)$  

*$(\sin30^\circ.\cos45^\circ + \sin45^\circ.\cos30^\circ) / (\cos30^\circ.\cos45^\circ - \sin45^\circ.\sin30^\circ)$   

*$\dfrac{(1/2\sqrt{2}) + (3/2\sqrt{2})}{(3/2\sqrt{2}) - (1/2\sqrt{2})}$  

*$\dfrac{(1 + \sqrt{3})}{(\sqrt{3}) - 1}$  

*multiply throughout by $(\sqrt{3}) + 1)$


Another alternative approach:


*

*$\tan75^\circ$   

*$\tan(45^\circ + 30^\circ)$  

*$\dfrac{\tan45^\circ + \tan30^\circ}{1-\tan45^\circ.\tan30^\circ}$  

*$\dfrac{1 + 1/\sqrt{3}}{1-1/\sqrt{3}}$  

*at point 6 in above alternative 


$$
\begin{array}{l}
\tan (75^{\circ})=\frac{\sin (75^{\circ})}{\cos (75^{\circ})}=\frac{\cos (15^{\circ})}{\sin (15^{\circ})}=\frac{\cos (30^{\circ} / 2)}{\sin (30^{\circ} / 2)}=\sqrt{\frac{1+\cos (30^{\circ})}{1-\cos (30^{\circ})}} \\
=\sqrt{\frac{1+\frac{\sqrt{3}}{2}}{1-\frac{\sqrt{3}}{2}}}=\sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}}=\sqrt{\frac{(2+\sqrt{3})^{2}}{(2-\sqrt{3})(2+\sqrt{3})}}=2+\sqrt{3}
\end{array}
$$
