The point [tex]$\left(\frac{5}{13}, y\right)$[/tex] in the fourth quadrant corresponds to angle [tex]$\theta$[/tex] on the unit circle.

[tex]\[
\begin{array}{l}
\sec \theta = \square \\
\cot \theta = \square
\end{array}
\][/tex]



Answer :

Sure, let's solve the given problem step-by-step:

1. Identify the [tex]\(x\)[/tex]-coordinate:
The [tex]\(x\)[/tex]-coordinate in the given point is [tex]\(\frac{5}{13}\)[/tex].

2. Determine the [tex]\(y\)[/tex]-coordinate:
Since the point [tex]\((\frac{5}{13}, y)\)[/tex] lies on the unit circle, it must satisfy the equation [tex]\(x^2 + y^2 = 1\)[/tex], i.e., [tex]\(\left(\frac{5}{13}\right)^2 + y^2 = 1\)[/tex].

Calculate [tex]\(\left(\frac{5}{13}\right)^2\)[/tex]:
[tex]\[ \left(\frac{5}{13}\right)^2 = \frac{25}{169} \][/tex]

Substitute back into the equation:
[tex]\[ \frac{25}{169} + y^2 = 1 \][/tex]

Solve for [tex]\(y^2\)[/tex]:
[tex]\[ y^2 = 1 - \frac{25}{169} \][/tex]
[tex]\[ y^2 = \frac{169}{169} - \frac{25}{169} \][/tex]
[tex]\[ y^2 = \frac{144}{169} \][/tex]

Take the square root of both sides and note that since the point is in the fourth quadrant, [tex]\(y\)[/tex] will be negative:
[tex]\[ y = -\sqrt{\frac{144}{169}} = -\frac{12}{13} \][/tex]

3. Determine [tex]\(\sec \theta\)[/tex]:
The secant function is defined as the reciprocal of the cosine function, which in this context is the [tex]\(x\)[/tex]-coordinate.
[tex]\[ \sec \theta = \frac{1}{\cos \theta} = \frac{1}{\frac{5}{13}} = \frac{13}{5} = 2.6 \][/tex]

4. Determine [tex]\(\cot \theta\)[/tex]:
The cotangent function is defined as the cosine of the angle divided by the sine of the angle. With [tex]\(\cos \theta = \frac{5}{13}\)[/tex] and [tex]\(\sin \theta = y = -\frac{12}{13}\)[/tex],
[tex]\[ \cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{\frac{5}{13}}{-\frac{12}{13}} = \frac{5}{-12} = -\frac{5}{12} \approx -0.4166666666666667 \][/tex]

Hence, the final results are:
[tex]\[ \sec \theta = 2.6 \][/tex]
[tex]\[ \cot \theta = -0.4166666666666667 \][/tex]

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