(tg\alpha + ctg\alpha)^2 = (\frac{1}{sin^2\alpha}) * cos^2\alpha zapis ??
obliczam lewą stronę
(tg\alpha+ctg\alpha)^2=(\frac{sin\alpha}{cos\alpha}+\frac{cos\alpha}{sin\alpha})^2=(\frac{sin\alpha*sin\alpha+cos\alpha*cos \alpha}{cos\alpha*sin\alpha})^2=
(\frac{sin^2\alpha+cos^2\alpha}{sin\alpha*cos\alpha)})^2=(\frac{1}{sin\alpha*cos\alpha})^2=\frac{1}{sin^2\alpha*cos^2\alpha}