Answer:
A
Step-by-step explanation:
We are given:
[tex]\displaystyle \cos(\theta)=-\frac{12}{13},\,\theta\in\text{QIII}[/tex]
Since cosine is the ratio of the adjacent side over the hypotenuse, this means that the opposite side is (we can ignore negatives for now):
[tex]o=\sqrt{13^2-12^2}=\sqrt{25}=5[/tex]
So, the opposite side is 5, the adjacent side is 12, and the hypotenuse is 13.
And since θ is in QIII, sine/cosecant is negative, cosine/secant is negative, and tangent/cotangent is positive.
Cosecant is given by the hypotenuse over the opposite side. Thus:
[tex]\displaystyle \csc(\theta)=\frac{13}{5}[/tex]
Since θ is in QIII, cosecant must be negative:
[tex]\displaystyle \csc(\theta)=-\frac{13}{5}[/tex]
Our answer is A.
