Нужна помощь с задачей номер 49

Решите задачу:

$ctg\Big (arccos(-\frac{1}{3})-\pi \Big )=- \frac{\sqrt2}{4}\\\\\\arccos(-\frac{1}{3})=\pi -arccos\frac{1}{3} \; \; \to \\\\\\ctg\Big (arccos(-\frac{1}{3})-\pi \Big )=ctg\Big (\pi -arccos\frac{1}{3}-\pi \Big )=ctg(-arccos\frac{1}{3})=\\\\=-ctg(arccos\frac{1}{3})\; ;\\\\\\ \alpha =arccos\frac{1}{3}\; \; \to \; \; cos\alpha =\frac{1}{3}\\\\sin^2 \alpha =1-cos^2 \alpha =1-\frac{1}{9}= \frac{8}{9} \\\\1+ctg^2 \alpha = \frac{1}{sin^2 \alpha } \; \; \to \; \; ctg^2 \alpha =\frac{1}{sin^2\alpha }-1= \frac{9}{8}-1=\frac{1}{8}$

$ctg \alpha =\pm \frac{1}{\sqrt8}=\pm \frac{1}{2\sqrt2}=\pm \frac{\sqrt2}{4} \\\\Tak\; kak\; \; cos \alpha = \frac{1}{3}\ \textgreater \ 0\; ,\; \; to\; \; \alpha =arccos\frac{1}{3}\in 1\; chetverti \; \; i\; \; ctg \alpha \ \textgreater \ 0\; ,\\\\ctg \alpha =ctg(arccos\frac{1}{3})=+\frac{\sqrt2}{4}\\\\\\ctg\Big (arccos(-\frac{1}{3})-\pi \Big )=-ctg(arccos\frac{1}{3})=-\frac{\sqrt2}{4}$