Okay, let's walk through solving the given question step-by-step.
We begin with the information provided:
[tex]\[
\cos \theta = \frac{15}{17}
\][/tex]
### Step 1: Compute [tex]\(\sec \theta\)[/tex]
We know that [tex]\(\sec \theta\)[/tex] is the reciprocal of [tex]\(\cos \theta\)[/tex]:
[tex]\[
\sec \theta = \frac{1}{\cos \theta} = \frac{1}{\frac{15}{17}} = \frac{17}{15}
\][/tex]
This matches option A, so A is correct.
### Step 2: Compute [tex]\(\sin \theta\)[/tex]
Next, we use the Pythagorean identity:
[tex]\[
\sin^2 \theta + \cos^2 \theta = 1
\][/tex]
Since [tex]\(\cos \theta = \frac{15}{17}\)[/tex], we have:
[tex]\[
\cos^2 \theta = \left(\frac{15}{17}\right)^2 = \frac{225}{289}
\][/tex]
Thus,
[tex]\[
\sin^2 \theta = 1 - \cos^2 \theta = 1 - \frac{225}{289} = \frac{289}{289} - \frac{225}{289} = \frac{64}{289}
\][/tex]
Taking the positive square root (since we're typically dealing with principal values unless otherwise specified):
[tex]\[
\sin \theta = \sqrt{\frac{64}{289}} = \frac{8}{17}
\][/tex]
Option B states [tex]\( \sin \theta = \frac{15}{8} \)[/tex]. Since [tex]\(\frac{8}{17} \neq \frac{15}{8}\)[/tex], this option is incorrect.
### Step 3: Compute [tex]\(\tan \theta\)[/tex]
We know:
[tex]\[
\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{8}{17}}{\frac{15}{17}} = \frac{8}{15}
\][/tex]
This matches option C, so C is correct.
### Step 4: Compute [tex]\(\csc \theta\)[/tex]
We know that [tex]\(\csc \theta\)[/tex] is the reciprocal of [tex]\(\sin \theta\)[/tex]:
[tex]\[
\csc \theta = \frac{1}{\sin \theta} = \frac{1}{\frac{8}{17}} = \frac{17}{8}
\][/tex]
Option D states that [tex]\(\csc \theta = \frac{17}{15}\)[/tex]. Since [tex]\(\frac{17}{8} \neq \frac{17}{15}\)[/tex], this option is incorrect.
### Summary of Valid Choices:
A. [tex]\(\sec \theta = \frac{17}{15}\)[/tex] is correct.
B. [tex]\(\sin \theta = \frac{15}{8}\)[/tex] is incorrect.
C. [tex]\(\tan \theta = \frac{8}{15}\)[/tex] is correct.
D. [tex]\(\csc \theta = \frac{17}{15}\)[/tex] is incorrect.
Thus, the correct options are:
[tex]\[
\boxed{A \text{ and } C}
\][/tex]
