keeping in mind that, on the II Quadrant, the sine/opposite is positive whilst the cosine/adjacent is negative, then
[tex]\bf sin(\theta )=\cfrac{\stackrel{opposite}{12}}{\stackrel{hypotenuse}{13}}\impliedby \textit{let's find the \underline{adjacent side}}
\\\\\\
\textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a
\qquad
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}
\\\\\\
\pm\sqrt{13^2-12^2}=a\implies \pm\sqrt{25}=a
\\\\\\
\pm 5=a\implies \stackrel{II~Quadrant}{-5=a}\qquad therefore\qquad
sec(\theta )=\cfrac{\stackrel{hypotenuse}{13}}{\stackrel{adjacent}{-5}}[/tex]
