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.
![(x,y)=(-(5)/(13),(12)/(13))](https://img.qammunity.org/2023/formulas/mathematics/college/s8vooi3jx7cmwzv0c38rtmi588k3fcl19r.png)
Let us find the given trigonometric ratios in the simplest rational form.
First, we need to find the hypotenuse side using the Pythagorean theorem.
![\begin{gathered} c^2=a^2+b^2 \\ c^2=(-(5)/(13))^2+((12)/(13))^2 \\ c^2=(25)/(169)+(144)/(169) \\ c^2=(169)/(169) \\ c^=\sqrt{(169)/(169)} \\ c=1 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/q31olz7rlt9ayt3itcgu2f8jy2jnro78du.png)
So, the hypotenuse side is 1.
![\begin{gathered} \tan\theta=(y)/(x) \\ \cos\theta=(x)/(h) \\ \csc\theta=(h)/(y) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/rq8xawwzcizramb9tzbxjepr9m459zlkam.png)
Where
x = -5/13
y = 12/13
h = 1
![\begin{gathered} \tan\theta=((12)/(13))/(-(5)/(13))=-(12)/(5) \\ \cos\theta=((-5)/(13))/(1)=-(5)/(13) \\ \csc\theta=(1)/((12)/(13))=(13)/(12) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/jx5atbzw2ujce6hhacquzfbqnwb4eoyq9e.png)
Therefore, the trigonometric ratios in the simplest rational form are
![\begin{gathered} \tan\theta=-(12)/(5) \\ \cos\theta=-(5)/(13) \\ \csc\theta=(13)/(12) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/duljjkm4oawv65p642or80xxo91p9qrflk.png)