Tg^2(arccos(-1/4)) помогите решить

Question

Tg^2(arccos(-1/4)) помогите решить

Answer 1

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

Answer 2

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