To find the values of [tex]\(\csc \theta\)[/tex] and [tex]\(\tan \theta\)[/tex] given that [tex]\(\cos \theta = -\frac{12}{13}\)[/tex] and [tex]\(\theta\)[/tex] is in quadrant III, we will follow these steps.
1. Determine [tex]\(\sin \theta\)[/tex]:
Since [tex]\(\theta\)[/tex] lies in quadrant III, both sine and cosine values are negative. We use the Pythagorean identity:
[tex]\[
\sin^2 \theta + \cos^2 \theta = 1
\][/tex]
Given [tex]\(\cos \theta = -\frac{12}{13}\)[/tex], we substitute and solve for [tex]\(\sin^2 \theta\)[/tex]:
[tex]\[
\sin^2 \theta + \left(-\frac{12}{13}\right)^2 = 1
\][/tex]
[tex]\[
\sin^2 \theta + \frac{144}{169} = 1
\][/tex]
[tex]\[
\sin^2 \theta = 1 - \frac{144}{169}
\][/tex]
[tex]\[
\sin^2 \theta = \frac{169}{169} - \frac{144}{169}
\][/tex]
[tex]\[
\sin^2 \theta = \frac{25}{169}
\][/tex]
Therefore,
[tex]\[
\sin \theta = -\sqrt{\frac{25}{169}}
\][/tex]
Since [tex]\(\theta\)[/tex] is in quadrant III (where sine is negative),
[tex]\[
\sin \theta = -\frac{5}{13}
\][/tex]
2. Find [tex]\(\csc \theta\)[/tex]:
The cosecant function is the reciprocal of the sine function:
[tex]\[
\csc \theta = \frac{1}{\sin \theta}
\][/tex]
Substituting the value of [tex]\(\sin \theta\)[/tex],
[tex]\[
\csc \theta = \frac{1}{-\frac{5}{13}} = -\frac{13}{5}
\][/tex]
3. Find [tex]\(\tan \theta\)[/tex]:
The tangent function is the ratio of sine to cosine:
[tex]\[
\tan \theta = \frac{\sin \theta}{\cos \theta}
\][/tex]
Substituting the values of [tex]\(\sin \theta\)[/tex] and [tex]\(\cos \theta\)[/tex],
[tex]\[
\tan \theta = \frac{-\frac{5}{13}}{-\frac{12}{13}} = \frac{-5}{-12} = \frac{5}{12}
\][/tex]
So, the exact values are:
[tex]\[
\csc \theta = -\frac{13}{5}
\][/tex]
[tex]\[
\tan \theta = \frac{5}{12}
\][/tex]
