Нужна помощь с задачей номер 35

Решите задачу:

$cos(arcctg(-\frac{1}{5}))=-\frac{\sqrt{26}}{26}\\\\\\ \alpha =arcctg(-\frac{1}{5})\; \; \; \Rightarrow \; \; \; ctg\alpha =-\frac{1}{5} \; \; \; \Rightarrow \; \; \; tg\alpha = \frac{1}{ctg \alpha } =-5\\\\\frac{1}{cos^2 \alpha} =1+tg^2\alpha \; \; \Rightarrow \; \; cos^2\alpha =\frac{1}{1+tg^2\alpha }= \frac{1}{1+(-5)^2} = \frac{1}{26}\; \; \Rightarrow \\\\cos\alpha =\pm \frac{1}{\sqrt{26}} \\\\Tak\; kak\; \; ctg\alpha =-\frac{1}{5}\ \textless \ 0,\; to\; \; \alpha =arcctg(-\frac{1}{5})\in 2\; chetverti\; \; i\; \; cos\alpha <0\\\\cos\alpha =-\frac{1}{\sqrt{26}}=-\frac{\sqrt{26}}{26}$