Question

АлгебрSin690°ctg1200°-1/2tg750°

Answer 1

Sin690*ctg1200-1/2*tg750=sin(2*360-30)*ctg(180*7-60)-1/2tg(180*4+30)=
=-sin30*(-ctg60)-1/2*tg30=-1/2*(-√3/3)-1/2*√3/3=√3/6-√3/6=0

Answer 2

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

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