Odpowiedź:
Szczegółowe wyjaśnienie:
Korzystamy z tożsamości trygonometrycznych:
[tex]\sin^2\alpha+\cos^2\alpha=1\ ,\qquad \text{tg\,}\alpha=\dfrac{\sin\alpha}{\cos\alpha}\ ,\qquad \text{ctg\,}\alpha=\dfrac{1}{\text{tg\,}\alpha}[/tex]
[tex]\sin^2\alpha+\cos^2\alpha=1\\\\\sin^2\alpha+(-\frac{24}{25})^2=1\\\\\sin^2\alpha+\frac{576}{625}=1\\\\\sin^2\alpha=\frac{49}{625}\quad\wedge\quad \alpha\in(0^o,180^o)\\\\\sin\alpha=\frac7{25}\\\\ \\\text{tg\,}\alpha=\dfrac{\frac7{25}}{-\frac{24}{25}}=-\frac7{25}\cdot\frac{25}{24}=-\frac7{24}\\\\ \\\text{ctg\,}\alpha=\dfrac{1}{-\frac7{24}}=-\frac{24}7=-3\frac37[/tex]
[tex]\sin^2\alpha+\cos^2\alpha=1\\\\(\frac{\sqrt5}{3})^2+\cos^2\alpha=1\\\\ \frac{5}{9}+\cos^2\alpha=1\\\\\cos^2\alpha=\frac{4}{9}\quad\wedge\quad \alpha\in(0^o,180^o)\\\\\cos\alpha=\frac23\qquad\vee\qquad\cos\alpha=- \frac23\\\\ \\\text{tg\,}\alpha=\dfrac{\frac{\sqrt5}3}{\frac23}=\frac{\sqrt5}2\qquad\vee\qquad\text{tg\,}\alpha=\dfrac{\frac{\sqrt5}3}{-\frac23}=-\frac{\sqrt5}2[/tex]
[tex]\text{ctg\,}\alpha=\dfrac{1}{\frac{\sqrt5}2}=\frac2{\sqrt5}=\frac{2\sqrt5}5\quad\vee\quad\text{ctg\,}\alpha=\dfrac{1}{-\frac{\sqrt5}2}=-\frac{2\sqrt5}5[/tex]
