Question

помогите пожалуйста​

Answer 1

$sina=-\frac{5}{\sqrt{41}};a\in \left ( 270^{\circ};360^{\circ} \right )\\a=(-1)^{k+1}\cdot arcsin\frac{5}{\sqrt{41}}+2\pi k\\a\in \left ( 270^{\circ};360^{\circ} \right )\Rightarrow a=-arcsin\frac{5}{\sqrt{41}}+2\pi$

Answer 2

$sina=-\dfrac{5}{\sqrt{41}}\\\\a\in (270^\circ ;360^\circ )\; \; \Rightarrow \; \; \; cosa>0\; ,\; \; tga<0\; ,\; \; ctga<0\\\\\\\boxed {\; sin^2a+cos^2a=1\; }\; \; \; \Rightarrow \; \; \; cos^2a=1-sin^2a\; ,\; \; cosa=\pm \sqrt{1-sin^2a}\\\\\\cosa>0\; \; \to \; \; \; cosa=+\sqrt{1-sin^2a}=\sqrt{1-\dfrac{25}{41}}=\sqrt{\dfrac{16}{41}}=\dfrac{4}{\sqrt{41}}\\\\\\tga=\dfrac{sina}{cosa}=\dfrac{-\frac{5}{\sqrt{41}}}{\frac{4}{\sqrt{41}}}=-\dfrac{5}{4}\\\\\\ctga=\dfrac{1}{tga}=-\dfrac{4}{5}$