It is given that the terminal side of angle θ passes through the point (-5/13, 12/13) in quadrant II on a unit circle.
[tex](x,y)=(-\frac{5}{13},\frac{12}{13})[/tex]
Let us find the given trigonometric ratios in the simplest rational form.
First, we need to find the hypotenuse side using the Pythagorean theorem.
[tex]\begin{gathered} c^2=a^2+b^2 \\ c^2=(-\frac{5}{13})^2+(\frac{12}{13})^2 \\ c^2=\frac{25}{169}+\frac{144}{169} \\ c^2=\frac{169}{169} \\ c^=\sqrt{\frac{169}{169}} \\ c=1 \end{gathered}[/tex]
So, the hypotenuse side is 1.
[tex]\begin{gathered} \tan\theta=\frac{y}{x} \\ \cos\theta=\frac{x}{h} \\ \csc\theta=\frac{h}{y} \end{gathered}[/tex]
Where
x = -5/13
y = 12/13
h = 1
[tex]\begin{gathered} \tan\theta=\frac{\frac{12}{13}}{-\frac{5}{13}}=-\frac{12}{5} \\ \cos\theta=\frac{\frac{-5}{13}}{1}=-\frac{5}{13} \\ \csc\theta=\frac{1}{\frac{12}{13}}=\frac{13}{12} \end{gathered}[/tex]
Therefore, the trigonometric ratios in the simplest rational form are
[tex]\begin{gathered} \tan\theta=-\frac{12}{5} \\ \cos\theta=-\frac{5}{13} \\ \csc\theta=\frac{13}{12} \end{gathered}[/tex]
