Answer :
Sure, let's break down the problem step by step.
Given:
- The point [tex]\(\left(\frac{5}{13}, y\right)\)[/tex] is in the fourth quadrant on the unit circle.
First, we know that the coordinates of any point [tex]\((x, y)\)[/tex] on the unit circle satisfy the equation [tex]\(x^2 + y^2 = 1\)[/tex].
1. Identify [tex]\( x = \frac{5}{13} \)[/tex].
Next, solve for [tex]\( y \)[/tex] using the unit circle equation:
[tex]\[ \left( \frac{5}{13} \right)^2 + y^2 = 1 \][/tex]
[tex]\[ \left( \frac{25}{169} \right) + y^2 = 1 \][/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]
[tex]\[ y = \pm \frac{\sqrt{144}}{\sqrt{169}} \][/tex]
[tex]\[ y = \pm \frac{12}{13} \][/tex]
Since the point is in the fourth quadrant, [tex]\(y\)[/tex] is negative:
[tex]\[ y = -\frac{12}{13} \][/tex]
Now we need to find [tex]\(\sec \theta\)[/tex] and [tex]\(\cot \theta\)[/tex]:
2. Compute [tex]\(\sec \theta\)[/tex]:
[tex]\[ \sec \theta = \frac{1}{\cos \theta} = \frac{1}{x} = \frac{1}{\frac{5}{13}} = \frac{13}{5} \][/tex]
Therefore, [tex]\(\sec \theta = 2.6\)[/tex].
3. Compute [tex]\(\cot \theta\)[/tex]:
[tex]\[ \cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{x}{y} = \frac{\frac{5}{13}}{-\frac{12}{13}} = \frac{5}{-12} = - \frac{5}{12} \][/tex]
Therefore, [tex]\(\cot \theta = -0.4166666666666667\)[/tex].
So the answers are:
[tex]\[ \sec \theta = 2.6 \][/tex]
[tex]\[ \cot \theta = -0.4166666666666667 \][/tex]
Given:
- The point [tex]\(\left(\frac{5}{13}, y\right)\)[/tex] is in the fourth quadrant on the unit circle.
First, we know that the coordinates of any point [tex]\((x, y)\)[/tex] on the unit circle satisfy the equation [tex]\(x^2 + y^2 = 1\)[/tex].
1. Identify [tex]\( x = \frac{5}{13} \)[/tex].
Next, solve for [tex]\( y \)[/tex] using the unit circle equation:
[tex]\[ \left( \frac{5}{13} \right)^2 + y^2 = 1 \][/tex]
[tex]\[ \left( \frac{25}{169} \right) + y^2 = 1 \][/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]
[tex]\[ y = \pm \frac{\sqrt{144}}{\sqrt{169}} \][/tex]
[tex]\[ y = \pm \frac{12}{13} \][/tex]
Since the point is in the fourth quadrant, [tex]\(y\)[/tex] is negative:
[tex]\[ y = -\frac{12}{13} \][/tex]
Now we need to find [tex]\(\sec \theta\)[/tex] and [tex]\(\cot \theta\)[/tex]:
2. Compute [tex]\(\sec \theta\)[/tex]:
[tex]\[ \sec \theta = \frac{1}{\cos \theta} = \frac{1}{x} = \frac{1}{\frac{5}{13}} = \frac{13}{5} \][/tex]
Therefore, [tex]\(\sec \theta = 2.6\)[/tex].
3. Compute [tex]\(\cot \theta\)[/tex]:
[tex]\[ \cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{x}{y} = \frac{\frac{5}{13}}{-\frac{12}{13}} = \frac{5}{-12} = - \frac{5}{12} \][/tex]
Therefore, [tex]\(\cot \theta = -0.4166666666666667\)[/tex].
So the answers are:
[tex]\[ \sec \theta = 2.6 \][/tex]
[tex]\[ \cot \theta = -0.4166666666666667 \][/tex]
