To solve this problem, we start with [tex]\(\sec \theta = -\frac{37}{12}\)[/tex] and aim to find [tex]\(\cot \theta\)[/tex] in the interval [tex]\(\frac{\pi}{2} < \theta < \pi\)[/tex].
1. Calculate [tex]\(\cos \theta\)[/tex]:
Since [tex]\(\sec \theta = \frac{1}{\cos \theta}\)[/tex], we can find [tex]\(\cos \theta\)[/tex] by taking the reciprocal of [tex]\(\sec \theta\)[/tex]. Therefore,
[tex]\[
\cos \theta = \frac{1}{\sec \theta} = \frac{1}{-\frac{37}{12}} = -\frac{12}{37}.
\][/tex]
2. Determine [tex]\(\sin \theta\)[/tex]:
We need to find [tex]\(\sin \theta\)[/tex] using the Pythagorean identity: [tex]\(\sin^2 \theta + \cos^2 \theta = 1\)[/tex].
Plugging in our value for [tex]\(\cos \theta\)[/tex],
[tex]\[
\cos^2 \theta = \left(-\frac{12}{37}\right)^2 = \frac{144}{1369}.
\][/tex]
Hence,
[tex]\[
\sin^2 \theta = 1 - \cos^2 \theta = 1 - \frac{144}{1369} = \frac{1369}{1369} - \frac{144}{1369} = \frac{1225}{1369}.
\][/tex]
Since [tex]\(\theta\)[/tex] is in the second quadrant ([tex]\(\frac{\pi}{2} < \theta < \pi\)[/tex]), where sine is positive, we take the positive square root:
[tex]\[
\sin \theta = \sqrt{\frac{1225}{1369}} = \frac{35}{37}.
\][/tex]
3. Calculate [tex]\(\cot \theta\)[/tex]:
We know that [tex]\(\cot \theta = \frac{\cos \theta}{\sin \theta}\)[/tex]. Using the values obtained,
[tex]\[
\cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{-\frac{12}{37}}{\frac{35}{37}} = -\frac{12}{35}.
\][/tex]
Therefore, the value of [tex]\(\cot \theta\)[/tex] is:
[tex]\[
-\frac{12}{35}.
\][/tex]
The correct choice is [tex]\(\boxed{-\frac{12}{35}}\)[/tex].
