Given the point [tex]\(\left(\frac{5}{13}, y\right)\)[/tex] on the unit circle in the fourth quadrant, let's determine the values of [tex]\(\sec \theta\)[/tex] and [tex]\(\cot \theta\)[/tex].
1. Find [tex]\(\sec \theta\)[/tex]:
- The x-coordinate of the point is [tex]\(\frac{5}{13}\)[/tex].
- Since secant is the reciprocal of cosine, [tex]\(\sec \theta = \frac{1}{\cos \theta}\)[/tex].
- Here, [tex]\(\cos \theta = \frac{5}{13}\)[/tex].
- Therefore, [tex]\(\sec \theta = \frac{1}{\frac{5}{13}} = \frac{13}{5} \approx 2.6\)[/tex].
So, [tex]\(\sec \theta \approx 2.6\)[/tex].
2. Find [tex]\(\cot \theta\)[/tex]:
- In the fourth quadrant, cosine is positive and sine is negative.
- To find [tex]\(\cot \theta\)[/tex], we use the relationship [tex]\(\cot \theta = \frac{\cos \theta}{\sin \theta}\)[/tex].
- From the given point, [tex]\(\cos \theta = \frac{5}{13}\)[/tex].
- We need to find the corresponding [tex]\(y\)[/tex]-coordinate, [tex]\(\sin \theta\)[/tex]. In the fourth quadrant, [tex]\(\sin \theta\)[/tex] is negative.
- We use the identity [tex]\(\cos^2 \theta + \sin^2 \theta = 1\)[/tex]:
[tex]\[
\left( \frac{5}{13} \right)^2 + \sin^2 \theta = 1
\][/tex]
[tex]\[
\frac{25}{169} + \sin^2 \theta = 1
\][/tex]
[tex]\[
\sin^2 \theta = 1 - \frac{25}{169} = \frac{144}{169}
\][/tex]
[tex]\[
\sin \theta = -\sqrt{\frac{144}{169}} = -\frac{12}{13}
\][/tex]
- Now, [tex]\(\cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{\frac{5}{13}}{-\frac{12}{13}} = -\frac{5}{12} \approx -0.417\)[/tex].
So, [tex]\(\cot \theta \approx -0.417\)[/tex].
Hence, the correct answers are:
[tex]\[
\begin{array}{l}
\sec \theta = 2.6 \\
\cot \theta = -0.417
\end{array}
\][/tex]
