[tex]\bf (\stackrel{a}{-4}~,~\stackrel{b}{7})\impliedby \textit{let's find the \underline{hypotenuse}}
\\\\\\
\textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies c=\sqrt{a^2+b^2}
\qquad
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}
\\\\\\
c=\sqrt{(-4)^2+7^2}\implies c=\sqrt{65}[/tex]
keeping in mind that, even though the square root would give us two roots, the positive and negative ones, the hypotenuse is however only a radius distance, and therefore is never negative.
[tex]\bf sin(\theta)=\cfrac{opposite}{hypotenuse}
\qquad\qquad
cos(\theta)=\cfrac{adjacent}{hypotenuse}
\\\\\\
% tangent
tan(\theta)=\cfrac{opposite}{adjacent}
\qquad \qquad
% cotangent
cot(\theta)=\cfrac{adjacent}{opposite}
\\\\\\
% cosecant
csc(\theta)=\cfrac{hypotenuse}{opposite}
\qquad \qquad
% secant
sec(\theta)=\cfrac{hypotenuse}{adjacent}\\\\
-------------------------------[/tex]
[tex]\bf sin(\theta )=\cfrac{7}{\sqrt{65}}\quad \stackrel{rationalized}{\implies }\quad\cfrac{7\sqrt{65}}{65}
\\\\\\
cos(\theta )=\cfrac{-4}{\sqrt{65}}\quad \stackrel{rationalized}{\implies }\quad \cfrac{-4\sqrt{65}}{65}
\\\\\\
tan(\theta )=-\cfrac{7}{4}\qquad\qquad \qquad cot(\theta )=-\cfrac{4}{7}
\\\\\\
csc(\theta )=\cfrac{\sqrt{65}}{7}\qquad\qquad \qquad sec(\theta )=-\cfrac{\sqrt{65}}{4}[/tex]
