Obliczenia
[tex]\sqrt{\frac{4cos^230^o+tg30^o\cdot tg60^o}{sin^233^o+sin^257^o}}+tg45^o=\sqrt{\frac{4\cdot(\frac{\sqrt3}{2})^2+\frac{\sqrt3}{3}\cdot\sqrt3}{sin^233^o+sin^2(90^o-33^o)}}+1=\\\\\\=\sqrt{\frac{4\cdot\frac{3}{4}+\frac{3}{3}}{sin^233^o+cos^233^o}}+1=\sqrt{\frac{3+1}{1}}+1=\sqrt4+1=2+1=\huge\boxed3[/tex]
Użyte wzory
[tex]sin(90^o-\alpha)=cos\alpha\\\\sin^2\alpha+cos^2\alpha=1[/tex]
