[tex]sin150^o \cdot tg225^o + sin^2300^o= sin(180^o-30^o) \cdot tg(180^o+45^o)+ \\ + sin^2(360^o-60^o) = sin30^o \cdot tg45^o + (-sin60^o)^2=\frac{1}{2} \cdot 1 + (-\frac{\sqrt{3}}{2})^2 = \\ = \frac{1}{2}+\frac{3}{4}= \frac{2}{4}+\frac{3}{4}= \frac{5}{4}=1\frac{1}{4}[/tex]
