Помогите решить. Напишите решение задания!

$1)\; \; sin\alpha=\frac{1}{\sqrt3} \; ,\\\\\frac{\pi}{2}\ \textless \ \alpha \ \textless \ \pi \; \; \Rightarrow\; \; cos \alpha \ \textless \ 0\; ,\; tg \alpha \ \textless \ 0\\\\cos \alpha =-\sqrt{1-sin^2 \alpha }=-\sqrt{1-\frac{1}{3}}=-\sqrt{\frac{2}{3}}=-\frac{\sqrt6}{3}\\\\tg \alpha =-\sqrt{\frac{1}{cos^2 \alpha }-1}=-\sqrt{\frac{3}{2}-1}=-\frac{1}{\sqrt2}=-\frac{\sqrt2}{2}\\\\2)\; \; cos \alpha =\frac{12}{13}\; ,\\\\\frac{3\pi}{2}\ \textless \ \alpha \ \textless \ 2\pi \; \; \Rightarrow \; \; sin \alpha \ \textless \ 0\; ,\; ctg \alpha \ \textless \ 0$

$sin \alpha =-\sqrt{1-cos^2 \alpha }=$ $-\sqrt{1-\frac{144}{169}}=-\sqrt{\frac{25}{169}}=-\frac{5}{13}$

$ctg \alpha =-\sqrt{\frac{1}{sin^2 \alpha }-1}=-\sqrt{\frac{169}{25}-1}=-\sqrt{\frac{144}{25}}=-\frac{12}{5}$