Question

1) 1-1/sin^2L
2)4-tgL*ctgL
3)cos^2b-cos^2bsin^2b

Answer 1

$1-\frac{1}{sin^2a}=\frac{sin^2a}{sin^2a}-\frac{1}{sin^2a}=\frac{sin^2a-1}{sin62a}=\frac{-(1-sin^2a)}{sin^2a}=\frac{-cos^2a}{sin^2a}=-ctg^2a\\\\4-tga*ctga=4-\frac{sina}{cosa}*\frac{cosa}{sina}= 4-1=3\\\\cos^2\beta-cos^2\beta*sin^2\beta=cos^2\beta(1-sin^2\beta)=cos^2\beta*cos^2\beta=cos^4\beta$