Question

АлгебрSin690°ctg1200°-1/2tg750°

Answer 1

$sin690^circ cdot ctg1200^circ - frac{1}{2} cdot tg750^circ =\\=sin(720^circ -30^circ)cdot ctg(4cdot 360^circ -240^circ)- frac{1}{2}cdot tg(720^circ+30^circ)=\\=sin(-30^circ)cdot ctg(-240^circ )- frac{1}{2} cdot tg30^circ=\\=-sin30^circ cdot (-ctg(360^circ-120^circ))- frac{1}{2}cdot frac{sqrt3}{3} =\\=-frac{1}{2}cdot ctg(180^circ-60^circ)- frac{sqrt3}{6}=-frac{1}{2}cdot (-ctg60^circ )- frac{sqrt3}{6}=$

$=frac{1}{2} cdot frac{sqrt3}{3}-frac{sqrt3}{6}=frac{sqrt3}{6}-frac{sqrt3}{6}=0$