Let's analyze the problem step-by-step:
Given: [tex]\(\csc \theta = \frac{13}{12}\)[/tex]
Step 1: Find [tex]\(\sin \theta\)[/tex]
[tex]\(\csc \theta\)[/tex] is the reciprocal of [tex]\(\sin \theta\)[/tex]. Therefore,
[tex]\[
\sin \theta = \frac{1}{\csc \theta} = \frac{1}{\frac{13}{12}} = \frac{12}{13}
\][/tex]
- A. [tex]\(\sin \theta = \frac{12}{13}\)[/tex] is True.
Step 2: Find [tex]\(\cos \theta\)[/tex] using the Pythagorean identity
The Pythagorean identity states:
[tex]\[
\sin^2 \theta + \cos^2 \theta = 1
\][/tex]
Substitute [tex]\(\sin \theta = \frac{12}{13}\)[/tex]:
[tex]\[
\left( \frac{12}{13} \right)^2 + \cos^2 \theta = 1 \\
\frac{144}{169} + \cos^2 \theta = 1 \\
\cos^2 \theta = 1 - \frac{144}{169} \\
\cos^2 \theta = \frac{169}{169} - \frac{144}{169} \\
\cos^2 \theta = \frac{25}{169}
\][/tex]
Taking the positive square root (since cosine can be positive in the first quadrant),
[tex]\[
\cos \theta = \frac{5}{13}
\][/tex]
- D. [tex]\(\cos \theta = \frac{12}{13}\)[/tex] is False.
Step 3: Find [tex]\(\tan \theta\)[/tex]
[tex]\(\tan \theta = \frac{\sin \theta}{\cos \theta}\)[/tex]:
[tex]\[
\tan \theta = \frac{\frac{12}{13}}{\frac{5}{13}} = \frac{12}{5}
\][/tex]
- B. [tex]\(\tan \theta = \frac{12}{5}\)[/tex] is False.
Step 4: Find [tex]\(\sec \theta\)[/tex]
[tex]\(\sec \theta\)[/tex] is the reciprocal of [tex]\(\cos \theta\)[/tex]:
[tex]\[
\sec \theta = \frac{1}{\cos \theta} = \frac{1}{\frac{5}{13}} = \frac{13}{5}
\][/tex]
- C. [tex]\(\sec \theta = \frac{12}{13}\)[/tex] is False.
Summary:
The correct answers are:
- A. [tex]\(\sin \theta = \frac{12}{13}\)[/tex] is True.
- B. [tex]\(\tan \theta = \frac{12}{5}\)[/tex] is False.
- C. [tex]\(\sec \theta = \frac{12}{13}\)[/tex] is False.
- D. [tex]\(\cos \theta = \frac{12}{13}\)[/tex] is False.
