Question

a) cos(-120)+sin1020+tg(-510)=?

б) ctg258+ctg(-303)/1+ctg(-237)*tg192=?

Answer 1

Решение Вашего задания во вложении

Answer 2

$1) cos(-120)+sin(1020)+tg(-510)=\\=cos120+sin(3*360-60)-tg510=\\=cos(180-60)+sin60-tg(360+150)=\\=-cos60+sin60-tg150=cos60+sin60-tg(180-30)=\\=cos60+sin60-tg30= \frac{1}{2} + \frac{ \sqrt{3} }{2} - \frac{ \sqrt{3} }{3} = \frac{1+ \sqrt{3} }{2} - \frac{ \sqrt{3} }{3} = \\ = \frac{3-3 \sqrt{3} -2 \sqrt{3} }{6} = \frac{3- \sqrt{3} }{6}$


$2) \frac{ctg258+ctg(-303)}{1+ctg(-237)*tg192} \\ \\ctg258+ctg(-303)=ctg258-ctg303=ctg(180+78)- \\ -ctg(360-57)= ctg78-ctg57=ctg(90-12)- \\ -ctg(90-33)=tg12-tg33 \\ \\ 1+ctg(-237)*tg192=1-ctg237*tg192= \\ =1-ctg(360-123)*tg(180+12)= \\ =1-ctg123*tg12=1-ctg(90+33)*tg12= \\ =1+tg33*tg12\\\\\\ \frac{tg12-tg33}{1+tg33*tg12} =tg(12+33)=tg45=1$