Question

Решить:
4cos^2(31pi/6)+tg^2(11pi/3)+ctg^2(29pi/4)

Answer 1

$4cos^2, frac{31pi}{6}+tg^2, frac{11pi}{3}+ctg^2, frac{29pi}{4}=\\=4cos^2(4pi +pi +frac{pi}{6})+tg^2(4pi -frac{pi}{3})+ctg^2(7pi +frac{pi}{4})=\\=4cos^2(pi +frac{pi}{6})+tg^2(-frac{pi}{3})+ctg^2frac{pi}{4}=4(-cosfrac{pi}{6})^2+(-tgfrac{pi}{3})^2+1^2=\\=4cos^2frac{pi}{6}+tg^2frac{pi}{3}+1=4cdot (frac{sqrt3}{2})^2+(sqrt3)^2+1=\\=4cdot frac{3}{4}+3+1=7$