Question

va rog ajutatima plssssssss

Answer 1

   
[tex]\displaystyle\\ \bf\\ ctg\;30^o = tg\;60^o ~~~\texttt{Deoarece } 30^o + 60^o = 90^o\\\\ \Big( tg \;60^o + sin \;60^o - ctg\;30\Big)^{-1} = \\\\ =\Big( tg \;60^o + sin \;60^o - tg\;60\Big)^{-1} = \\\\ =\Big( sin \;60^o \Big)^{-1} = \Big( \frac{ \sqrt{3}}{2}\Big)^{-1}=\frac{ 2}{ \sqrt{3}} = \boxed{\bf \frac{ 2\sqrt{3}}{ 3} }[/tex]