Упр.669 ГДЗ Никольский Потапов 9 класс (Алгебра)

б) (tg72°-tg12°)/(1+tg39°•tg6°)
в) tg5°
г) tg15°
д) tg105

Воспользуемся формулами:

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

1. $$\frac{\tg 39^\circ+\tg 6^\circ}{1-\tg 39^\circ\cdot\tg 6^\circ}=\tg(39^\circ+6^\circ)=\tg 45^\circ=1.$$

2. $$\frac{\tg 72^\circ-\tg 12^\circ}{1+\tg 72^\circ\cdot\tg 12^\circ}=\tg(72^\circ-12^\circ)=\tg 60^\circ=\sqrt{3}.$$

3. $$\tg 75^\circ=\tg(45^\circ+30^\circ)=\frac{\tg 45^\circ+\tg 30^\circ}{1-\tg 45^\circ\cdot\tg 30^\circ}=\frac{1+\frac{1}{\sqrt{3}}}{1-\frac{1}{\sqrt{3}}}.$$

   $$\frac{1+\frac{1}{\sqrt{3}}}{1-\frac{1}{\sqrt{3}}}=\frac{\sqrt{3}+1}{\sqrt{3}-1}=\frac{(\sqrt{3}+1)^2}{(\sqrt{3}-1)(\sqrt{3}+1)}=\frac{3+2\sqrt{3}+1}{3-1}=2+\sqrt{3}.$$

4. $$\tg 15^\circ=\tg(45^\circ-30^\circ)=\frac{\tg 45^\circ-\tg 30^\circ}{1+\tg 45^\circ\cdot\tg 30^\circ}=\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}.$$

   $$\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}=\frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{(\sqrt{3}-1)^2}{(\sqrt{3}-1)(\sqrt{3}+1)}=\frac{3-2\sqrt{3}+1}{3-1}=2-\sqrt{3}.$$

5. $$\tg 105^\circ=\tg(60^\circ+45^\circ)=\frac{\tg 60^\circ+\tg 45^\circ}{1-\tg 60^\circ\cdot\tg 45^\circ}=\frac{\sqrt{3}+1}{1-\sqrt{3}}.$$

   $$\frac{\sqrt{3}+1}{1-\sqrt{3}}=\frac{(\sqrt{3}+1)(1+\sqrt{3})}{(1-\sqrt{3})(1+\sqrt{3})}=\frac{4+2\sqrt{3}}{-2}=-2-\sqrt{3}.$$

Ответ

а) $$1$$; б) $$\sqrt{3}$$; в) $$2+\sqrt{3}$$; г) $$2-\sqrt{3}$$; д) $$-2-\sqrt{3}$$.