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

\(\frac{sin^4 α - cos^4 α}{(1-cos α)(1+cos α)} + 2ctg^2 α = \frac{(sin^2 α - cos^2 α)(sin^2 α + cos^2 α)}{1 - cos^2 α} + 2ctg^2 α = \frac{(sin^2 α - cos^2 α) * 1}{sin^2 α} + 2ctg^2 α = \frac{sin^2 α - cos^2 α}{sin^2 α} + 2 \frac{cos^2 α}{sin^2 α} = 1 - \frac{cos^2 α}{sin^2 α} + 2 \frac{cos^2 α}{sin^2 α} = 1 + \frac{cos^2 α}{sin^2 α} = \frac{sin^2 α + cos^2 α}{sin^2 α} = \frac{1}{sin^2 α}\) Тождество верно.