1) Если $\sin \alpha = \frac{2\sqrt{29}}{29}$ и $\alpha \in (0; \frac{\pi}{2})$, то $\cos \alpha$ будет положительным.
Используем основное тригонометрическое тождество $\sin^2 \alpha + \cos^2 \alpha = 1$:
$$\cos^2 \alpha = 1 - \sin^2 \alpha = 1 - \left(\frac{2\sqrt{29}}{29}\right)^2 = 1 - \frac{4 \cdot 29}{29^2} = 1 - \frac{4}{29} = \frac{25}{29}$$
$$\cos \alpha = \sqrt{\frac{25}{29}} = \frac{5}{\sqrt{29}} = \frac{5\sqrt{29}}{29}$$
Теперь найдём $\text{tg} \alpha$:
$$\text{tg} \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{\frac{2\sqrt{29}}{29}}{\frac{5\sqrt{29}}{29}} = \frac{2\sqrt{29}}{29} \cdot \frac{29}{5\sqrt{29}} = \frac{2}{5}$$
**Ответ: $\frac{2}{5}$**
2) Если $\sin \alpha = \frac{\sqrt{26}}{26}$ и $\alpha \in (0; \frac{\pi}{2})$, то $\cos \alpha$ будет положительным.
$$\cos^2 \alpha = 1 - \sin^2 \alpha = 1 - \left(\frac{\sqrt{26}}{26}\right)^2 = 1 - \frac{26}{26^2} = 1 - \frac{1}{26} = \frac{25}{26}$$
$$\cos \alpha = \sqrt{\frac{25}{26}} = \frac{5}{\sqrt{26}} = \frac{5\sqrt{26}}{26}$$
$$\text{tg} \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{\frac{\sqrt{26}}{26}}{\frac{5\sqrt{26}}{26}} = \frac{\sqrt{26}}{26} \cdot \frac{26}{5\sqrt{26}} = \frac{1}{5}$$
**Ответ: $\frac{1}{5}$**
3) Если $\cos \alpha = \frac{\sqrt{17}}{17}$ и $\alpha \in (0; \frac{\pi}{2})$, то $\sin \alpha$ будет положительным.
$$\sin^2 \alpha = 1 - \cos^2 \alpha = 1 - \left(\frac{\sqrt{17}}{17}\right)^2 = 1 - \frac{17}{17^2} = 1 - \frac{1}{17} = \frac{16}{17}$$
$$\sin \alpha = \sqrt{\frac{16}{17}} = \frac{4}{\sqrt{17}} = \frac{4\sqrt{17}}{17}$$
$$\text{tg} \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{\frac{4\sqrt{17}}{17}}{\frac{\sqrt{17}}{17}} = \frac{4\sqrt{17}}{17} \cdot \frac{17}{\sqrt{17}} = 4$$
**Ответ: $4$**
4) Если $\cos \alpha = \frac{\sqrt{5}}{5}$ и $\alpha \in (0; \frac{\pi}{2})$, то $\sin \alpha$ будет положительным.
$$\sin^2 \alpha = 1 - \cos^2 \alpha = 1 - \left(\frac{\sqrt{5}}{5}\right)^2 = 1 - \frac{5}{25} = 1 - \frac{1}{5} = \frac{4}{5}$$
$$\sin \alpha = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{5}$$
$$\text{tg} \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{\frac{2\sqrt{5}}{5}}{\frac{\sqrt{5}}{5}} = \frac{2\sqrt{5}}{5} \cdot \frac{5}{\sqrt{5}} = 2$$
**Ответ: $2$**
5) Если $\sin \alpha = \frac{3\sqrt{34}}{34}$ и $\alpha \in (\frac{\pi}{2}; \pi)$, то $\cos \alpha$ будет отрицательным.
$$\cos^2 \alpha = 1 - \sin^2 \alpha = 1 - \left(\frac{3\sqrt{34}}{34}\right)^2 = 1 - \frac{9 \cdot 34}{34^2} = 1 - \frac{9}{34} = \frac{25}{34}$$
$$\cos \alpha = -\sqrt{\frac{25}{34}} = -\frac{5}{\sqrt{34}} = -\frac{5\sqrt{34}}{34}$$
$$\text{tg} \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{\frac{3\sqrt{34}}{34}}{-\frac{5\sqrt{34}}{34}} = \frac{3\sqrt{34}}{34} \cdot \left(-\frac{34}{5\sqrt{34}}\right) = -\frac{3}{5}$$
**Ответ: $-\frac{3}{5}$**
6) Если $\sin \alpha = \frac{2\sqrt{5}}{5}$ и $\alpha \in (\frac{\pi}{2}; \pi)$, то $\cos \alpha$ будет отрицательным.
$$\cos^2 \alpha = 1 - \sin^2 \alpha = 1 - \left(\frac{2\sqrt{5}}{5}\right)^2 = 1 - \frac{4 \cdot 5}{25} = 1 - \frac{20}{25} = 1 - \frac{4}{5} = \frac{1}{5}$$
$$\cos \alpha = -\sqrt{\frac{1}{5}} = -\frac{1}{\sqrt{5}} = -\frac{\sqrt{5}}{5}$$
$$\text{tg} \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{\frac{2\sqrt{5}}{5}}{-\frac{\sqrt{5}}{5}} = \frac{2\sqrt{5}}{5} \cdot \left(-\frac{5}{\sqrt{5}}\right) = -2$$
**Ответ: $-2$**
7) Если $\cos \alpha = -\frac{\sqrt{10}}{10}$ и $\alpha \in (\frac{\pi}{2}; \pi)$, то $\sin \alpha$ будет положительным.
$$\sin^2 \alpha = 1 - \cos^2 \alpha = 1 - \left(-\frac{\sqrt{10}}{10}\right)^2 = 1 - \frac{10}{100} = 1 - \frac{1}{10} = \frac{9}{10}$$
$$\sin \alpha = \sqrt{\frac{9}{10}} = \frac{3}{\sqrt{10}} = \frac{3\sqrt{10}}{10}$$
$$\text{tg} \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{\frac{3\sqrt{10}}{10}}{-\frac{\sqrt{10}}{10}} = \frac{3\sqrt{10}}{10} \cdot \left(-\frac{10}{\sqrt{10}}\right) = -3$$
**Ответ: $-3$**
8) Если $\cos \alpha = -\frac{\sqrt{26}}{26}$ и $\alpha \in (\frac{\pi}{2}; \pi)$, то $\sin \alpha$ будет положительным.
$$\sin^2 \alpha = 1 - \cos^2 \alpha = 1 - \left(-\frac{\sqrt{26}}{26}\right)^2 = 1 - \frac{26}{26^2} = 1 - \frac{1}{26} = \frac{25}{26}$$
$$\sin \alpha = \sqrt{\frac{25}{26}} = \frac{5}{\sqrt{26}} = \frac{5\sqrt{26}}{26}$$
$$\text{tg} \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{\frac{5\sqrt{26}}{26}}{-\frac{\sqrt{26}}{26}} = \frac{5\sqrt{26}}{26} \cdot \left(-\frac{26}{\sqrt{26}}\right) = -5$$
**Ответ: $-5$**
