Ctg-tg/ctg+tg

Решите задачу:

$ctg\frac{\pi}{12}-tg\frac{\pi}{12} = \frac{cos\frac{\pi}{12}}{sin\frac{\pi}{12}}-\frac{sin\frac{\pi}{12}}{cos\frac{\pi}{12}}=\frac{cos^2\frac{\pi}{12}-sin^2\frac{\pi}{12}}{sin\frac{\pi}{12}cos\frac{\pi}{12}} = \frac{cos\frac{2\pi}{12}}{\frac{1}{2}sin\frac{2\pi}{12}}=\\=2\frac{cos\frac{\pi}{6}}{sin\frac{\pi}{6}}=2^*\frac{\frac{\sqrt3}{2}}{\frac{1}{2}}=2\sqrt3$
$ctg\frac{\pi}{8}+tg\frac{\pi}{8} = \frac{cos\frac{\pi}{8}}{sin\frac{\pi}{8}}+\frac{sin\frac{\pi}{8}}{cos\frac{\pi}{8}}=\frac{cos^2\frac{\pi}{8}+sin^2\frac{\pi}{8}}{cos\frac{\pi}{8}sin\frac{\pi}{8}}=\frac{\frac{1+cos\frac{2\pi}{8}}{2}+\frac{1-cos\frac{2\pi}{8}}{2}}{\frac{1}{2}sin(2^*\frac{\pi}{8})}=\\=\frac{2}{sin\frac{\pi}{4}}=\frac{2}{\frac{\sqrt2}{2}}=2\sqrt2$
$\frac{2\sqrt3}{2\sqrt2}=\frac{\sqrt6}{2}$