Страница 407 номер 1300, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение
Упростить выражение (1297—1302). 1) \(\frac{\operatorname{tg}^2 \alpha}{1 + \operatorname{ctg}^2 \alpha}\); 2) \(\frac{1 + \operatorname{ctg}^2 \alpha}{\operatorname{ctg}^2 \alpha}\); 3) \(\frac{\operatorname{tg} \alpha - \operatorname{tg} \beta}{\operatorname{ctg} \alpha + \operatorname{ctg} \beta}\); 4) \((\operatorname{tg} \alpha + \operatorname{ctg} \alpha)^2 - (\operatorname{tg} \alpha - \operatorname{ctg} \alpha)^2\).
1) \(\frac{\operatorname{tg}^{2}\alpha}{1 + \operatorname{ctg}^{2}\alpha} = \frac{\operatorname{tg}^{2}\alpha}{1 + \frac{1}{\operatorname{tg}^{2}\alpha}} = \frac{\operatorname{tg}^{4}\alpha}{1 + \operatorname{tg}^{2}\alpha} = \frac{\sin^{4}\alpha}{\cos^{4}\alpha} \cdot \frac{\cos^{2}\alpha}{1} = \sin^{2}\alpha\operatorname{tg}^{2}\alpha\);
2) \(\frac{1 + \operatorname{ctg}^{2}\alpha}{\operatorname{ctg}^{2}\alpha} = \frac{1}{\operatorname{ctg}^{2}\alpha} + 1 = \operatorname{tg}^{2}\alpha + 1 = \frac{1}{\cos^{2}\alpha}\);
3) \(\frac{\operatorname{tg} \alpha - \operatorname{tg} \beta}{\operatorname{ctg} \alpha + \operatorname{ctg} \beta} = \frac{\frac{\sin\alpha}{\cos\alpha} - \frac{\sin\beta}{\cos\beta}}{\frac{\cos\alpha}{\sin\alpha} + \frac{\cos\beta}{\sin\beta}} = \frac{\sin\alpha\cos\beta - \sin\beta\cos\alpha}{\cos\alpha\cos\beta} \cdot \frac{\sin\alpha\sin\beta}{\sin\beta\cos\alpha + \cos\beta\sin\alpha} =\)
\(= \operatorname{tg} \alpha \cdot \operatorname{tg} \beta \cdot \dfrac{\sin(\alpha - \beta)}{\sin(\alpha + \beta)};\)
4) \(\left( \operatorname{tg} \alpha + \operatorname{ctg} \alpha \right)^{2} - \left( \operatorname{tg} \alpha - \operatorname{ctg} \alpha \right)^{2} =\)
\(= \operatorname{tg}^{2}\alpha + 2 + \operatorname{ctg}^{2}\alpha - \operatorname{tg}^{2}\alpha + 2 - \operatorname{ctg}^{2}\alpha = 2 + 2 = 4.\)
