now, keeping in mind that, sine is opposite/hypotenuse, and that the hypotenuse is just a radius unit, and therefore is never negative, then, in the fraction of -3/5, the negative values has to be the numerator, not the hypotenuse.
something else to keep in mind is that, cosine > 0, meaning is positive, that only happens in the I and IV quadrants.
since we know the sine is negative, and the cosine is positive, the only place that occurs is on the IV quadrant, so then θ is in the IV quadrant.
[tex]\bf sin(\theta )=\cfrac{\stackrel{opposite}{-3}}{\stackrel{hypotenuse}{5}}\impliedby \textit{now 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{5^2-(-3)^2}=a\implies \pm\sqrt{25-9}=a\implies \pm\sqrt{16}=a
\\\\\\
\pm 4=a\implies \stackrel{IV~quadrant}{4=a}[/tex]
[tex]\bf cos(\theta)=\cfrac{adjacent}{hypotenuse}
\qquad
% tangent
tan(\theta)=\cfrac{opposite}{adjacent}
\\\\\\
% cotangent
cot(\theta)=\cfrac{adjacent}{opposite}
\qquad
% cosecant
csc(\theta)=\cfrac{hypotenuse}{opposite}
\quad
% secant
sec(\theta)=\cfrac{hypotenuse}{adjacent}\\\\
-------------------------------\\\\
cos(\theta)=\cfrac{4}{5}\qquad tan(\theta)=\cfrac{-3}{4}\qquad cot(\theta)=\cfrac{4}{-3}
\\\\\\
csc(\theta)=\cfrac{5}{-3}\qquad sec(\theta)=\cfrac{5}{4}[/tex]
