Question

Найдите cosα, если п<α<3п/2 и tgα=√3

Answer 1

$\displaystyle tga=\sqrt{3}; a \in III; cosa<0\\\\1+tg^2a=\frac{1}{cos^2a}\\\\cos^2a=\frac{1}{1+tg^2a}=\frac{1}{1+3}=\frac{1}{4}\\\\cosa=-\frac{1}{2}$