Ребят, помогите пожалуйста №79(3)и №80(2,3)

79(3) $\frac{sin^2 \alpha }{1+cos \alpha } = \frac{1-cos^2 \alpha }{1+cos \alpha} = \frac{(1-cos \alpha)(1+cos \alpha)}{1+cos \alpha} =1-cos \alpha$
80(2) $cos^2 \alpha +ctg^2 \alpha +sin^2 \alpha =1+ctg^2 \alpha = 1+\frac{cos^2 \alpha }{sin^2 \alpha } = \frac{sin^2 \alpha+cos^2 \alpha}{sin^2 \alpha} = \frac{1}{ sin^2 \alpha}$
при α=π/6
$\frac{1}{ sin^2 \frac{ \pi }{6}}=\frac{1}{ (\frac{ 1 }{2} )^2}= \frac{1}{ \frac{1}{4} }=4$
80(3) $\frac{1}{cos^2 \alpha } -1= \frac{1-cos^2 \alpha}{cos^2 \alpha} = \frac{sin^2 \alpha}{cos^2 \alpha} =tg^2 \alpha$
при α=π/3
$tg^2 \frac{ \pi }{3} = (\sqrt{3}) ^2=3$