[tex]\displaystyle\bf\\4sin\,60^o-\sqrt{3}cos\frac{\pi}{6}+ctg\frac{\pi}{4}=?\\\frac{\pi}{6}=\frac{180}{6}=30^o\\\\\frac{\pi}{4}=\frac{180}{4}=45^o\\\\4sin\,60^o-\sqrt{3}cos\,30^o+ctg\,45^o=?\\sin\,60^o=\frac{\sqrt{3}}{2}\\cos\,30^o=\frac{\sqrt{3}}{2}\\ctg\,45^o=1\\4sin\,60^o-\sqrt{3}cos\,30^o+ctg\,45^o=\\=4\times\frac{\sqrt{3}}{2}-\sqrt{3}\times\frac{\sqrt{3}}{2}+1=\\\\=\frac{4\sqrt{3}}{2}-\frac{\sqrt{3}\times\sqrt{3}}{2}+1=\\=2\sqrt{3}-\frac{3}{2}+1=2\sqrt{3}-1,\!5+1=\boxed{\bf2\sqrt{3}-0,\!5}[/tex]
