Question

Решите пожалуйста, задание 2

Answer 1

$cos alpha =frac{sqrt6}{3} textgreater 0; ,; alpha in (pi ;2pi ); ; Rightarrow ; ; tg alpha textless 0\\1+tg^2 alpha =frac{1}{cos^2 alpha }; ; Rightarrow ; ; tg^2 alpha =frac{1}{cos^2 alpha }-1\\sqrt2tg alpha =-sqrt2cdot sqrt{frac{1}{cos^2 alpha }-1}=-sqrt2cdot sqrt{frac{1}{6/9}-1}=-sqrt2cdot sqrt{frac{9}{6}-1}=\\=-sqrt2cdot sqrt{frac{3}{2}-1}=-sqrt2cdot sqrt{frac{1}{2}}=-sqrt2cdot frac{1}{sqrt2}=-1$