例 2 如图①，在$ Rt \triangle ABC$中，$\angle C = 90^{\circ}$，$\angle A = 15^{\circ}$，不用计算器，求$\tan 15^{\circ}$的值.

解法一:作点 C 关于 AB 的对称点 E，连接 EB，AE，并延长 AE 交 CB 的延长线于点 D(如图②)，则$\angle CAE = 30^{\circ}$，$\angle D = 60^{\circ}$.
令$ED = a$，$\therefore BD = 2a$，$CB = EB = \sqrt{3}a$，$\therefore AC = \sqrt{3}CD = \sqrt{3} × (2a + \sqrt{3}a)$.
$\therefore \tan 15^{\circ} = \tan \angle CAB = \frac{BC}{AC} = \frac{\sqrt{3}a}{\sqrt{3} × (2 + \sqrt{3})a} = 2 - \sqrt{3}$.
解法二:以 AC 为边在 AB 同侧作$\angle CAD = 30^{\circ}$，延长 CB 交 AD 于点 D，延长 AC 至点 F，使$AF = AD$，连接 DF(如图③)，则$\angle ADF = \angle AFD = 75^{\circ}$，
$\therefore \angle CDF = 15^{\circ}$.
令$CD = a$，$\therefore AC = \sqrt{3}a$，$AF = AD = 2a$，$CF = 2a - \sqrt{3}a$.
$\therefore \tan 15^{\circ} = \tan \angle CDF = \frac{CF}{CD} = 2 - \sqrt{3}$.
解法三:以 AC 为边在 AB 异侧作$\angle CAH = 30^{\circ}$，延长 BC 交 AH 于点 H，作$BG \perp AH$于点 G(如图④)，则$\angle HAB = 45^{\circ}$，$\angle AHB = 60^{\circ}$，$\therefore \angle GBH = 30^{\circ}$.
令$GH = a$，$\therefore BH = 2a$，$AG = GB = \sqrt{3}a$，$AH = a + \sqrt{3}a$，$CH = \frac{1}{2}(a + \sqrt{3}a)$.
$\therefore BC = 2a - \frac{1}{2}(a + \sqrt{3}a)$，$AC = \frac{\sqrt{3}}{2}(a + \sqrt{3}a)$.
$\therefore \tan 15^{\circ} = \tan \angle CAB = \frac{BC}{AC} = 2 - \sqrt{3}$.