3. Упростить выражения: 1) \( \frac{\operatorname{tg} \alpha + \operatorname{ctg} \beta}{\cos (\alpha - \beta)} \cdot \sin \alpha \sin \beta \); 2) \( \frac{1+\operatorname{tg} \alpha}{1-\operatorname{tg} \alpha} - \frac{1+\sin 2\alpha}{\cos 2\alpha} \)

Решение:

1. \( \frac{\operatorname{tg} \alpha + \operatorname{ctg} \beta}{\cos (\alpha - \beta)} \cdot \sin \alpha \sin \beta = \frac{\frac{\sin \alpha}{\cos \alpha} + \frac{\cos \beta}{\sin \beta}}{\cos \alpha \cos \beta + \sin \alpha \sin \beta} \cdot \sin \alpha \sin \beta \)


2. \( = \frac{\frac{\sin \alpha \sin \beta + \cos \alpha \cos \beta}{\cos \alpha \sin \beta}}{\cos \alpha \cos \beta + \sin \alpha \sin \beta} \cdot \sin \alpha \sin \beta = \frac{\cos(\alpha - \beta)}{\cos \alpha \sin \beta} \cdot \frac{\sin \alpha \sin \beta}{\cos(\alpha - \beta)} = \frac{\sin \alpha}{\cos \alpha} = \operatorname{tg} \alpha \)


3. \( \frac{1+\operatorname{tg} \alpha}{1-\operatorname{tg} \alpha} - \frac{1+\sin 2\alpha}{\cos 2\alpha} = \frac{1+\frac{\sin \alpha}{\cos \alpha}}{1-\frac{\sin \alpha}{\cos \alpha}} - \frac{1+2\sin \alpha \cos \alpha}{\cos^2 \alpha - \sin^2 \alpha} \)


4. \( = \frac{\frac{\cos \alpha + \sin \alpha}{\cos \alpha}}{\frac{\cos \alpha - \sin \alpha}{\cos \alpha}} - \frac{(\cos \alpha + \sin \alpha)^2}{(\cos \alpha - \sin \alpha)(\cos \alpha + \sin \alpha)} = \frac{\cos \alpha + \sin \alpha}{\cos \alpha - \sin \alpha} - \frac{\cos \alpha + \sin \alpha}{\cos \alpha - \sin \alpha} = 0 \)

Ответ: 1) \( \operatorname{tg} \alpha \); 2) 0.