1)Найти [tex]sin \alpha [/tex] и [tex]tg \alpha [/tex], если [tex]cos \alpha =- \frac{2}{5} [/tex] и [tex] \pi \ \textless \ \alpha \ \textless \ \frac{3 \pi }{2} [/tex]
2) Найти [tex]cos \alpha [/tex] и [tex]ctg \alpha [/tex], если [tex]sin \alpha = \frac{3}{4} [/tex]
и [tex] \frac{ \pi }{2} \ \textless \ \alpha \ \textless \ \pi [/tex]

[tex]1)\; \; cos \alpha =-\frac{2}{5}\; ,\\\\sin \alpha =\pm \sqrt{1-cos^2 \alpha }\; ,\; tg \alpha =\pm \sqrt{\frac{1}{cos^2 \alpha }-1}\\\\\; \; \ \alpha \in (\pi ;\frac{3\pi}{2})\; \; \to \; \; sin \alpha \ \textless \ 0,\; \; tg \alpha \ \textgreater \ 0\\\\sin \alpha =-\sqrt{1-\frac{4}{25}}=-\sqrt{\frac{21}{25}}=-\frac{\sqrt{21}}{5}\\\\tg \alpha =+\sqrt{\frac{25}{4}-1}=\frac{\sqrt{21}}{2}[/tex]

[tex]2)\; \; sin \alpha =\frac{3}{4}\\\\ \alpha \in (\frac{\pi}{2};\pi )\; \; \to \; \; cos \alpha \ \textless \ 0\; ,\; \; ctg \alpha \ \textless \ 0\\\\cos \alpha =-\sqrt{1-sin^2 \alpha }=-\sqrt{1-\frac{9}{16}}=-\frac{\sqrt7}{4}\\\\ctg \alpha =-\sqrt{\frac{1}{sin^2 \alpha }-1}=-\sqrt{\frac{16}{9}-1}=-\frac{\sqrt7}{3} [/tex]