Помогите пжл срочно.....

Решите задачу:

$1. \\ 1+tg^2x= \frac{1}{cos^2x} \\ cos2x = cos^2x-sin^2x \\sin^2x+cos^2x=1 =\ \textgreater \ (0.2)^2+cos^2x=1 \\ cos^2x=1-0.04 \\ cos^2x=0.96 \\ cos= ^+_- \sqrt{0.96} \\ 2.\\ \\ 1+tg^2x*cos2x= \frac{cos^2x-sin^2x}{cos^2x}= 1- \frac{sin^2x}{cos^2x}= \left \{ {{1- \frac{(0.2)^2}{ (\sqrt{0.96})^2}}=1- \frac{0.04}{0.96} \atop {1- \frac{(0.2)^2}{ (-\sqrt{0.96})^2}}=1- \frac{0.04}{0.96}}\right.= \\ = 1- \frac{1}{24} = \frac{23}{24}$