[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]
