Question

1).3tg 45*x tg 60*2).2cos 30*+6cos 60*-4tg 45*3).2ctg 60*-2sin 60*

Answer 1

3·tg 45°·tg 60° = 3·1·$\sqrt{3}$ = 3$\sqrt{3}$
2·cos 30°+ 6·cos 60° - 4·tg 45° = 2·$\frac{ \sqrt{3} }{2}$ + 6·$\frac{ 1}{2}$ - 4·1 = $\sqrt{3}+ 3 - 4 =\sqrt{3} - 1$
2·ctg 60° - 2·sin 60° = 2·$\frac{\sqrt{3} }{3}$ - 2·$\frac{ \sqrt{3} }{2}$ = 2$\sqrt{3}$·($\frac{1}{3}- \frac{1}{2}$) = $-\frac{\sqrt{3} }{3}$