tg135^{\circ}+cos 120^{\circ}\cdot sin 150^{\circ}=\\ = tg(180^{\circ}-45^{\circ})+cos(180^{\circ}-120^{\circ})\cdot sin(180^{\circ}-30^{\circ})\\=-tg45^{\circ}+(-cos60^{\circ})\cdot sin 35^{\circ}\\=-1-\frac{1}{2} \cdot \frac{1}{2}\\=-1-\frac{1}{4}\\ =-1\frac{1}{4}
Odpowiedź
-1\frac{1}{4}