$\tan75^o=\cot(90^o-75^o)=\cot15^o=2+\sqrt3$
$\to \tan15^o=\dfrac{1}{\cot15^o}=\dfrac{1}{2+\sqrt3}=\dfrac{2-\sqrt3}{4-3}=2-\sqrt3$
Ta có: $\dfrac{1}{\cos^215^o}=\dfrac{\sin^215^o+\cos^215^o}{\cos^215^o}=\tan^215^o+1$
$\to \cos^215^o=\dfrac{1}{1+(2-\sqrt3)^2}=\dfrac{2+\sqrt3}{4}$
$\to \cos15^o=\dfrac{\sqrt{2+\sqrt3}}{2}=\dfrac{\sqrt{4+2\sqrt3}}{2\sqrt2}=\dfrac{ \sqrt{ (\sqrt3+1)^2}}{2\sqrt2}=\dfrac{\sqrt3+1}{2\sqrt2}=\dfrac{\sqrt6+\sqrt2}{4}$
$\sin^215^o=1-\cos^215^o=\dfrac{2-\sqrt3}{4}$
$\to \sin15^o=\dfrac{\sqrt{2-\sqrt3}}{2}=\dfrac{\sqrt{4-2\sqrt3}}{2\sqrt2}=\dfrac{ \sqrt{ (\sqrt3-1)^2}}{2\sqrt2}=\dfrac{\sqrt3-1}{2\sqrt2}=\dfrac{\sqrt6-\sqrt2}{4}$
