The relation between polar and cartesian coordinates is given by:
In this case, we have:
[tex]P(x,y)=P(5,-12)\Rightarrow r={\sqrt{x^2+y^2}}=\sqrt{5^2+(-12)^2}=\sqrt{169}=13.[/tex]AnswerUsing the formulas above and the values of x, y and r, we have:
1) sin θ
[tex]\sinθ=\frac{y}{r}=-\frac{12}{13}.[/tex]2) cos θ
[tex]\cosθ=\frac{x}{r}=\frac{5}{13}.[/tex]3) tan θ
[tex]\tanθ=\frac{y}{x}=-\frac{12}{5}.[/tex]4) csc θ
[tex]csc\text{ }\theta=\frac{1}{sin\text{ }\theta}=\frac{1}{(-\frac{12}{13})}=-\frac{13}{12}.[/tex]5) sec θ
[tex]sec\text{ }\theta=\frac{1}{cos\text{ }\theta}=\frac{1}{\frac{5}{13}}=\frac{13}{5}.[/tex]6) cot θ
[tex]cot\text{ }\theta=\frac{1}{\tan\theta}=\frac{1}{(-\frac{12}{5})}=-\frac{5}{12}.[/tex]