Упростите выражения: 1) (tg α + ctg α) sin α cos α; 2) sin²α cos²α (tg²α + ctg²α + 2); 3) cos²α + tg²α cos²α; 4) sin²α + ctg²α sin²α; 5) (sin³α + cos³α)/(sin α + cos α) + sin α cos α; 6) (cos³α - sin³α)/(1 + sin α cos α) - cos α + sin α

**Ответ:** 1) $1$ 2) $1$ 3) $1$ 4) $1$ 5) $1$ 6) $0$ **Решение:** 1) $(\text{tg } \alpha + \text{ctg } \alpha) \sin \alpha \cos \alpha = (\frac{\sin \alpha}{\cos \alpha} + \frac{\cos \alpha}{\sin \alpha}) \sin \alpha \cos \alpha = \frac{\sin^2 \alpha + \cos^2 \alpha}{\sin \alpha \cos \alpha} \cdot \sin \alpha \cos \alpha = \sin^2 \alpha + \cos^2 \alpha = 1$ 2) $\sin^2 \alpha \cos^2 \alpha (\text{tg}^2 \alpha + \text{ctg}^2 \alpha + 2) = \sin^2 \alpha \cos^2 \alpha (\text{tg } \alpha + \text{ctg } \alpha)^2 = \sin^2 \alpha \cos^2 \alpha (\frac{\sin^2 \alpha + \cos^2 \alpha}{\sin \alpha \cos \alpha})^2 = \sin^2 \alpha \cos^2 \alpha \cdot \frac{1}{\sin^2 \alpha \cos^2 \alpha} = 1$ 3) $\cos^2 \alpha + \text{tg}^2 \alpha \cos^2 \alpha = \cos^2 \alpha (1 + \text{tg}^2 \alpha) = \cos^2 \alpha \cdot \frac{1}{\cos^2 \alpha} = 1$ 4) $\sin^2 \alpha + \text{ctg}^2 \alpha \sin^2 \alpha = \sin^2 \alpha (1 + \text{ctg}^2 \alpha) = \sin^2 \alpha \cdot \frac{1}{\sin^2 \alpha} = 1$ 5) $\frac{\sin^3 \alpha + \cos^3 \alpha}{\sin \alpha + \cos \alpha} + \sin \alpha \cos \alpha = \frac{(\sin \alpha + \cos \alpha)(\sin^2 \alpha - \sin \alpha \cos \alpha + \cos^2 \alpha)}{\sin \alpha + \cos \alpha} + \sin \alpha \cos \alpha = (1 - \sin \alpha \cos \alpha) + \sin \alpha \cos \alpha = 1$ 6) $\frac{\cos^3 \alpha - \sin^3 \alpha}{1 + \sin \alpha \cos \alpha} - \cos \alpha + \sin \alpha = \frac{(\cos \alpha - \sin \alpha)(\cos^2 \alpha + \cos \alpha \sin \alpha + \sin^2 \alpha)}{1 + \sin \alpha \cos \alpha} - \cos \alpha + \sin \alpha = \frac{(\cos \alpha - \sin \alpha)(1 + \sin \alpha \cos \alpha)}{1 + \sin \alpha \cos \alpha} - \cos \alpha + \sin \alpha = \cos \alpha - \sin \alpha - \cos \alpha + \sin \alpha = 0$