Question

Пожалуйста, помогите

Answer 1

$\frac{\pi }{2}<\alpha<\pi \\\\\alpha \in2chetverti\\\\Cos\alpha=-\frac{\sqrt{6} }{4}\\\\Sin\alpha>0\\\\Sin\alpha=\sqrt{1-Cos^{2}\alpha}=\sqrt{1-(-\frac{\sqrt{6}}{4})^{2}}=\sqrt{1-\frac{6}{16} }=\sqrt{\frac{10}{16} }=\frac{\sqrt{10}}{4}\\\\tg\alpha=\frac{Sin\alpha }{Cos\alpha}=\frac{\sqrt{10} }{4}:(-\frac{\sqrt{6} }{4})=- \frac{\sqrt{10} }{4}*\frac{4}{\sqrt{6}}=-\frac{\sqrt{5} }{\sqrt{3}}=-\frac{\sqrt{15} }{3}$

$Ctg\alpha=\frac{1}{tg\alpha}=-\frac{3}{\sqrt{15}}=-\frac{\sqrt{15} }{5}$