Решение:
- a) \( \sin(\pi + \alpha) + \cos\left(\frac{3\pi}{2} - \alpha\right) = -\sin \alpha + (-\sin \alpha) = -2\sin \alpha \).
- б) \( \text{tg}\left(\frac{\pi}{2} + \alpha\right) - \text{ctg}(2\pi - \alpha) = -\text{ctg} \alpha - (-\text{ctg} \alpha) = -\text{ctg} \alpha + \text{ctg} \alpha = 0 \).
- в) \( \cos 2\alpha + 2\sin^2(\pi - \alpha) = \cos 2\alpha + 2\sin^2 \alpha \).
- г) \( \frac{\sin \alpha}{1 + \cos \alpha} + \frac{\sin \alpha}{1 - \cos \alpha} = \frac{\sin \alpha (1 - \cos \alpha) + \sin \alpha (1 + \cos \alpha)}{(1 + \cos \alpha)(1 - \cos \alpha)} = \frac{\sin \alpha - \sin \alpha \cos \alpha + \sin \alpha + \sin \alpha \cos \alpha}{1 - \cos^2 \alpha} = \frac{2\sin \alpha}{\sin^2 \alpha} = \frac{2}{\sin \alpha} \).
Ответ: a) \(-2\sin \alpha\); б) 0; в) \(\cos 2\alpha + 2\sin^2 \alpha\); г) \(\frac{2}{\sin \alpha}\).