Ta có:
$\sin15^o=sin(60^o-45^o)$
$= \sin60^o.\cos 45^o-\cos 60^o.\sin45^o$
$=\dfrac{\sqrt3}{2}.\dfrac{\sqrt2}{2}-\dfrac{1}{2}\dfrac{\sqrt2}{2}$
$=\dfrac{\sqrt6}{4}-\dfrac{\sqrt2}{4}$
$=\dfrac{\sqrt2(\sqrt3-1)}{4}$
$\cos 15^o=\cos(60^o-45^o)$
$=\sin60^o.\cos45^o+\cos60^o.\sin45^o$
$=\dfrac{\sqrt3}{2}.\dfrac{\sqrt2}{2}+\dfrac{1}{2}.\dfrac{\sqrt2}{2}$
$=\dfrac{\sqrt6}{4}+\dfrac{\sqrt2}{4}$
$=\dfrac{\sqrt6+\sqrt2}{4}$
$=\dfrac{\sqrt2(\sqrt3+1)}{4}$
$\Rightarrow \tan 15^o=\dfrac{\sin 15^o}{\cos 15^o}=\dfrac{\sqrt3-1}{\sqrt3+1}=\dfrac{(\sqrt3-1)^2}{(\sqrt3-1)(\sqrt3+1)}=\dfrac{3-2\sqrt3+1}{3-1}=\dfrac{4-2\sqrt3}{2}=2-\sqrt3$
Vậy $\tan 15^o=2-\sqrt3$
