Страница 406 номер 1298, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение
Упростить выражение (1297—1302). 1) \(\frac{\operatorname{tg} \alpha + \operatorname{tg} \beta}{\operatorname{ctg} \alpha + \operatorname{ctg} \beta}\); 2) \((\sin \alpha + \cos \alpha)^2 + (\sin \alpha - \cos \alpha)^2\); 3) \(\frac{\sin\left(\frac{\pi}{4} + \alpha\right) - \cos\left(\frac{\pi}{4} + \alpha\right)}{\sin\left(\frac{\pi}{4} + \alpha\right) + \cos\left(\frac{\pi}{4} + \alpha\right)}\); 4) \(\frac{\sin \alpha + 2\sin\left(\frac{\pi}{3} - \alpha\right)}{2\cos\left(\frac{\pi}{6} - \alpha\right) - \sqrt{3}\cos \alpha}\).
1) \(\frac{\operatorname{tg} \alpha + \operatorname{tg} \beta}{\operatorname{ctg} \alpha + \operatorname{ctg} \beta} = \frac{\operatorname{tg} \alpha + \operatorname{tg} \beta}{\frac{1}{\operatorname{tg} \alpha} + \frac{1}{\operatorname{tg} \beta}} = \operatorname{tg} \alpha\operatorname{tg} \beta\);
2) \(\left( \sin\alpha + \cos\alpha \right)^{2} + \left( \sin\alpha - \cos\alpha \right)^{2} = 1 + \sin{2\alpha} + 1 - \sin{2\alpha} = 2\);
3) \(\frac{\sin\left( \frac{\pi}{4} + \alpha \right) - \cos\left( \frac{\pi}{4} + \alpha \right)}{\sin\left( \frac{\pi}{4} + \alpha \right) + \cos\left( \frac{\pi}{4} + \alpha \right)} = \frac{\sin\frac{\pi}{4}\cos\alpha + \cos\frac{\pi}{4}\sin\alpha - \cos\frac{\pi}{4}\cos\alpha + \sin\frac{\pi}{4}\sin\alpha}{\sin\frac{\pi}{4}\cos\alpha + \cos\frac{\pi}{4}\sin\alpha + \cos\frac{\pi}{4}\cos\alpha - \sin\frac{\pi}{4}\sin\alpha} =\)
\(= \dfrac{\cos\alpha + \sin\alpha - \cos\alpha + \sin\alpha}{\cos\alpha + \sin\alpha + \cos\alpha - \sin\alpha} = \dfrac{2\sin\alpha}{2\cos\alpha} = \operatorname{tg} \alpha;\)
4) \(\frac{\sin\alpha + 2\sin\left( \frac{\pi}{3} - \alpha \right)}{2\cos\left( \frac{\pi}{6} - \alpha \right) - \sqrt{3}\cos\alpha} = \frac{\sin\alpha + 2\sin\frac{\pi}{3}\cos\alpha - 2\cos\frac{\pi}{3}\sin\alpha}{2\cos\frac{\pi}{6}\cos\alpha + 2\sin\frac{\pi}{6}\sin\alpha - \sqrt{3}\cos\alpha} =\)
\(= \dfrac{\sin\alpha + \sqrt{3}\cos\alpha - \sin\alpha}{\sqrt{3}\cos\alpha + \sin\alpha - \sqrt{3}\cos\alpha} = \dfrac{\sqrt{3}\cos\alpha}{\sin\alpha} = \sqrt{3}\operatorname{ctg} \alpha.\)
