Упр.1438 ГДЗ Макарычев Миндюк 9 класс (Углубленный) (Алгебра)

а) (tg(20°)+tg(25°))/(1-tg(20°)tg(25°));
б) (tg(70°)-tg(10°))/(1+tg(70°)tg(10°));
в) (tg(7?/24)-tg(?/8))/(1+tg(7?/24)tg(?/8));
г) (tg(?/20)+tg(?/5))/(1-tg(?/20)tg(?/5)).

Используем формулы:

$$\frac{\tg \alpha+\tg \beta}{1-\tg \alpha \tg \beta}=\tg(\alpha+\beta),$$

$$\frac{\tg \alpha-\tg \beta}{1+\tg \alpha \tg \beta}=\tg(\alpha-\beta).$$

1. $$\frac{\tg 20^\circ+\tg 25^\circ}{1-\tg 20^\circ \tg 25^\circ}=\tg(20^\circ+25^\circ)=\tg 45^\circ=1.$$

2. $$\frac{\tg 70^\circ-\tg 10^\circ}{1+\tg 70^\circ \tg 10^\circ}=\tg(70^\circ-10^\circ)=\tg 60^\circ=\sqrt{3}.$$

3. $$\frac{\tg \frac{7\pi}{24}-\tg \frac{\pi}{8}}{1+\tg \frac{7\pi}{24}\tg \frac{\pi}{8}}=\tg\left(\frac{7\pi}{24}-\frac{\pi}{8}\right)=\tg\left(\frac{7\pi}{24}-\frac{3\pi}{24}\right)=\tg \frac{\pi}{6}=\frac{\sqrt{3}}{3}.$$

4. $$\frac{\tg \frac{\pi}{20}+\tg \frac{\pi}{5}}{1-\tg \frac{\pi}{20}\tg \frac{\pi}{5}}=\tg\left(\frac{\pi}{20}+\frac{\pi}{5}\right)=\tg\left(\frac{\pi}{20}+\frac{4\pi}{20}\right)=\tg \frac{\pi}{4}=1.$$

Ответ

а) $$1$$; б) $$\sqrt{3}$$; в) $$\frac{\sqrt{3}}{3}$$; г) $$1$$.