#### Answer

$tan~15^{\circ} = 2-\sqrt{3}$

#### Work Step by Step

Note that triangle $ACD$ is a right triangle since angle $ACD = 90^{\circ}$
We can use triangle $ACD$ to find $tan~15^{\circ}$:
$tan~15^{\circ} = \frac{opposite}{adjacent}$
$tan~15^{\circ} = \frac{AC}{CD}$
$tan~15^{\circ} = \frac{1}{2+\sqrt{3}}$
$tan~15^{\circ} = \frac{1}{2+\sqrt{3}}~\times \frac{2-\sqrt{3}}{2-\sqrt{3}}$
$tan~15^{\circ} = \frac{2-\sqrt{3}}{1}$
$tan~15^{\circ} = 2-\sqrt{3}$