Tg^2 (arccos (-1/4)) помогите решить

Надо привести к виду cos (arccos (x)) = x

давайте выведем формулу 1+tg^2=1/cos^2

tg^2 (arccos (-1/4)) = sin^2 (arccos (-1/4)) / cos^2 (arccos (-1/4)) = (1-cos^2 (arccos (-1/4))) / cos^2 (arccos (-1/4)) = (1 - (-1/4) * (-1/4)) / (-1/4*-1/4) = (1-1/16) / 1/16=15/16 / 1/16=15

Sin^2 (arccos (-1/4) / cos^2 (arccos (-1/4)) = (1-cos^2 (arccos (-1/4))) / cos^2 (arccos (-1/4)) =

= (1-1/16) / 1/16 = (15/16) / (1/16) = 15