Докажите тождества: \(\frac{cos^4 α - sin^4 α}{(1-sin α)(1 + sin α)} + 2tg^2 α = \frac{1}{cos^2 α}\) \(\frac{sin^4 α - cos^4 α}{(1-cos α)(1+cos α)} + 2 ctg^2 α = \frac{1}{sin^2 α}\)

1) \(\frac{cos^4 α - sin^4 α}{(1-sin α)(1 + sin α)} + 2tg^2 α = \frac{(cos^2 α - sin^2 α)(cos^2 α + sin^2 α)}{1 - sin^2 α} + 2tg^2 α = \frac{(cos^2 α - sin^2 α) \cdot 1}{cos^2 α} + 2tg^2 α = \frac{cos^2 α - sin^2 α}{cos^2 α} + 2 \cdot \frac{sin^2 α}{cos^2 α} = 1 - \frac{sin^2 α}{cos^2 α} + \frac{2sin^2 α}{cos^2 α} = 1 + \frac{sin^2 α}{cos^2 α} = 1 + tg^2 α = \frac{1}{cos^2 α}\) 2) \(\frac{sin^4 α - cos^4 α}{(1-cos α)(1+cos α)} + 2 ctg^2 α = \frac{(sin^2 α - cos^2 α)(sin^2 α + cos^2 α)}{1 - cos^2 α} + 2 ctg^2 α = \frac{(sin^2 α - cos^2 α) \cdot 1}{sin^2 α} + 2 ctg^2 α = \frac{sin^2 α - cos^2 α}{sin^2 α} + 2 \cdot \frac{cos^2 α}{sin^2 α} = 1 - \frac{cos^2 α}{sin^2 α} + \frac{2cos^2 α}{sin^2 α} = 1 + \frac{cos^2 α}{sin^2 α} = 1 + ctg^2 α = \frac{1}{sin^2 α}\)