(sin 11П/18 - sin П/18)/(cos 11П/18 - cos П/18)

Формулы, которые нам понадобятся:

$\boxed{\diplaystyle sin \alpha-sin \beta=2sin \frac{\alpha-\beta}{2} \cdot cos \frac{\alpha+\beta}{2}}$
$\boxed{\displaystyle cos \alpha - cos \beta =-2sin \frac{\alpha+\beta}{2} \cdot sin \frac{\alpha-\beta}{2}}$

$\displaystyle \frac{sin \frac{11 \pi}{18}-sin \frac{\pi}{18}}{cos \frac{11 \pi}{18}-cos \frac{\pi}{18}}=\frac{2sin \frac{\frac{11 \pi}{18}- \frac{\pi}{18}}{2} \cdot cos \frac{\frac{11 \pi}{18}+\frac{\pi}{18}}{2}}{-2sin \frac{\frac{11 \pi}{18}+ \frac{\pi}{18}}{2} \cdot sin \frac{\frac{11 \pi}{18}- \frac{\pi}{18}}{2}}=$
$\displaystyle = \frac{2sin (\frac{10 \pi}{18} \cdot \frac{1}{2}) \cdot cos (\frac{12 \pi}{18} \cdot \frac{1}{2})}{-2sin( \frac{12 \pi}{18} \cdot \frac{1}{2}) \cdot sin(\frac{10 \pi}{18}\cdot \frac{1}{2})}=\frac{2sin \frac{5 \pi}{18}\cdot cos \frac{\pi}{3}}{-2sin \frac{\pi}{3} \cdot sin \frac{5 \pi}{18}}=$
$\displaystyle = \frac{cos \frac{\pi}{3}}{-sin \frac{\pi}{3}}=-ctg \frac{\pi}{3}=\boxed{- \frac{\sqrt 3}{3}}$