Question

1)Ctg330° * sin135°
2)tg300°+3ctg240°

Answer 1

1)ctg330° × sin135°=ctg(360°-30°)×sin(90°+45°)=
=-ctg30°×cos45°=
$= - \sqrt{3} \times \frac{ \sqrt{2} }{2} = - \frac{ \sqrt{6} }{2}$
2)tg300°+3ctg240°=tg(360°-60°)+3ctg(270°-30°)=-tg60°+3tg30°=
$= - \sqrt{3} + 3 \frac{ \sqrt{3} }{3} = - \sqrt{3} + \sqrt{3} = 0$