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

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