Страница 148 номер 493, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение
Вычислить: 1) \(\frac{\operatorname{tg} 29^\circ + \operatorname{tg} 31^\circ}{1 - \operatorname{tg} 29^\circ \operatorname{tg} 31^\circ}\); 2) \(\frac{\operatorname{tg}\frac{7\pi}{16} - \operatorname{tg}\frac{3\pi}{16}}{1 + \operatorname{tg}\frac{7\pi}{16} \operatorname{tg}\frac{3\pi}{16}}\); 3) \(\frac{1 + \operatorname{tg} 10^\circ \operatorname{tg} 55^\circ}{\operatorname{tg} 55^\circ - \operatorname{tg} 10^\circ}\); 4) \(\frac{1 - \operatorname{tg} 13^\circ \operatorname{tg} 17^\circ}{\operatorname{tg} 17^\circ + \operatorname{tg} 13^\circ}\).
\(\operatorname{tg}(\alpha - \beta) = \operatorname{tg}(\alpha + (-\beta)) = \dfrac{\operatorname{tg} \alpha + \operatorname{tg}(-\beta)}{1 - \operatorname{tg} \alpha \operatorname{tg}(-\beta)} = \dfrac{\operatorname{tg} \alpha - \operatorname{tg} \beta}{1 + \operatorname{tg} \alpha \operatorname{tg} \beta}.\)
1) \(\frac{\operatorname{tg} 29^{\circ} + \operatorname{tg} 31^{\circ}}{1 - \operatorname{tg} 29^{\circ} \operatorname{tg} 31^{\circ}} = \operatorname{tg}(29^{\circ} + 31^{\circ}) = \operatorname{tg} 60^{\circ} = \sqrt{3}\);
2) \(\frac{\operatorname{tg} \frac{7\pi}{16} - \operatorname{tg} \frac{3\pi}{16}}{1 + \operatorname{tg} \frac{7\pi}{16} \operatorname{tg} \frac{3\pi}{16}} = \operatorname{tg}\left(\frac{7\pi}{16} - \frac{3\pi}{16}\right) = \operatorname{tg} \frac{\pi}{4} = 1\);
3) \(\frac{1 + \operatorname{tg} 10^{\circ} \operatorname{tg} 55^{\circ}}{\operatorname{tg} 55^{\circ} - \operatorname{tg} 10^{\circ}} = \frac{1}{\operatorname{tg}(55^{\circ} - 10^{\circ})} = \frac{1}{\operatorname{tg} 45^{\circ}} = 1\);
4) \(\frac{1 - \operatorname{tg} 13^{\circ} \operatorname{tg} 17^{\circ}}{\operatorname{tg} 17^{\circ} + \operatorname{tg} 13^{\circ}} = \frac{1}{\operatorname{tg}(17^{\circ} + 13^{\circ})} = \frac{1}{\operatorname{tg} 30^{\circ}} = \sqrt{3}\).
