Exercise 4.2.5 solution
\(x=-1,\quad y=2,\quad r=\sqrt{(-1)^2+2^2}=\sqrt{5}\)
\(\sin A =\dfrac{y}{r}=\dfrac{2}{\sqrt{5}}=\dfrac{2\sqrt{5}}{5},\quad \csc A=\dfrac{\sqrt{5}}{2}\)
\(\cos A =\dfrac{x}{r}=-\dfrac{1}{\sqrt{5}}=-\dfrac{\sqrt{5}}{5},\quad \sec A=-\sqrt{5}\)
\(\tan A=\dfrac{y}{x}=-2,\quad \cot A=-\dfrac{1}{2}\)