Question

помогите пожалуйста умоляю ​

Answer 1

$cosa=\frac{2}{\sqrt{5}}\Leftrightarrow a=\pm arccos\frac{2}{\sqrt{5}}+2\pi k,k\in \mathbb{Z}\\\left ( 270;360^{\circ}^{\circ} \right )\Rightarrow -arccos\frac{2}{\sqrt{5}}+2\pi$

Answer 2

$cosa=\frac{2}{\sqrt5}\\\\a\in (270^\circ ;360^\circ )\; \; \Rightarrow \; \; \; sina<0\; ,\; \; tga<0\; ,\; \; ctga<0\\\\\\\boxed {\; sin^2a+cos^2a=1\; }\; \; \; \Rightarrow \; \; \; sin^2a=1-cos^2a\; ,\; \; sina=\pm \sqrt{1-cos^2a}\\\\\\sina<0\; \; \to \; \; \; sina=-\sqrt{1-cos^2a}=-\sqrt{1-\dfrac{4}{5}}=-\sqrt{\dfrac{1}{5}}=-\dfrac{1}{\sqrt5}\\\\\\tga=\dfrac{sina}{cosa}=\dfrac{-\frac{1}{\sqrt5}}{\frac{2}{\sqrt5}}=-\dfrac{1}{2}\\\\\\ctga=\dfrac{1}{tga}=-2$