1115. 1) \( \frac{tg^2 a}{1+tg^2 a} \) 2) \( \frac{1+ctg^2 a}{ctg^2 a} \) 3) \( \frac{tg \alpha - tg \beta}{ctg \alpha + ctg \beta} \) 4) \( (tg a+ ctg a)^2-(tg a-ctg a)^2 \)

Решение:

1) \( \frac{tg^2 a}{1+tg^2 a} = \frac{\frac{\sin^2 a}{\cos^2 a}}{\frac{1}{\cos^2 a}} = \sin^2 a \)

2) \( \frac{1+ctg^2 a}{ctg^2 a} = \frac{1}{\frac{\cos^2 a}{\sin^2 a}} + 1 = \frac{\sin^2 a}{\cos^2 a} + 1 = \frac{\sin^2 a + \cos^2 a}{\cos^2 a} = \frac{1}{\cos^2 a} = sec^2 a \)

3) \( \frac{tg \alpha - tg \beta}{ctg \alpha + ctg \beta} = \frac{\frac{\sin \alpha}{\cos \alpha} - \frac{\sin \beta}{\cos \beta}}{\frac{\cos \alpha}{\sin \alpha} + \frac{\cos \beta}{\sin \beta}} = \frac{\frac{\sin \alpha \cos \beta - \cos \alpha \sin \beta}{\cos \alpha \cos \beta}}{\frac{\cos \alpha \sin \beta + \cos \beta \sin \alpha}{\sin \alpha \sin \beta}} = \frac{\sin(\alpha-\beta)}{\cos \alpha \cos \beta} \cdot \frac{\sin \alpha \sin \beta}{\sin(\alpha+\beta)} = \frac{\sin(\alpha-\beta) \tan \alpha \tan \beta}{\sin(\alpha+\beta)} \)

4) \( (tg a+ ctg a)^2-(tg a-ctg a)^2 = (tg^2 a + 2tg a ctg a + ctg^2 a) - (tg^2 a - 2tg a ctg a + ctg^2 a) = (tg^2 a + 2 + ctg^2 a) - (tg^2 a - 2 + ctg^2 a) = 4 tg a ctg a = 4 \)

Ответ: 1) \( \sin^2 a \); 2) \( sec^2 a \); 3) \( \frac{\sin(\alpha-\beta) \tan \alpha \tan \beta}{\sin(\alpha+\beta)} \); 4) 4