Упр.13.26 ГДЗ Мордковича 10 класс профильный уровень (Алгебра)

$$\frac{\sin \frac{\pi}{4}-\cos \pi-\tg \frac{\pi}{4}}{2\sin \frac{\pi}{6}-\sin \frac{3\pi}{2}}=\frac{\frac{\sqrt2}{2}-(-1)-1}{2\cdot \frac12-(-1)}=\frac{\frac{\sqrt2}{2}}{2}=\frac{\sqrt2}{4}.$$

$$\frac{\ctg \frac{5\pi}{4}+\sin \frac{3\pi}{2}\cdot \tg\!\left(-\frac{5\pi}{4}\right)}{2\cos \frac{11\pi}{6}+2\sin^2 \frac{11\pi}{4}}=\frac{\ctg\!\left(\frac{5\pi}{4}-\pi\right)-\sin \frac{3\pi}{2}\cdot \tg \frac{5\pi}{4}}{2\cos\!\left(\frac{11\pi}{6}-2\pi\right)+2\sin^2\!\left(2\pi-\frac{11\pi}{4}\right)}$$

$$=\frac{\ctg \frac{\pi}{4}-(-1)\cdot \tg\!\left(\frac{5\pi}{4}-\pi\right)}{2\cos\!\left(-\frac{\pi}{6}\right)+2\sin^2\!\left(-\frac{3\pi}{4}\right)}=\frac{1+1\cdot \tg \frac{\pi}{4}}{2\cdot \frac{\sqrt3}{2}+2\cdot \left(\frac{\sqrt2}{2}\right)^2}$$

$$=\frac{1+1}{\sqrt3+1}=\frac{2}{\sqrt3+1}=\frac{2(\sqrt3-1)}{(\sqrt3+1)(\sqrt3-1)}=\frac{2(\sqrt3-1)}{3-1}=\sqrt3-1.$$

Ответ: а) $$\frac{\sqrt2}{4}$$; б) $$\sqrt3-1$$.