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 


You can rather use $\tan (75)=\tan(45+30)$ and plug into the formula by Metin. Cause: Your $15^\circ$ is not so trivial.
