Certainly! Let’s solve for [tex]\(\sin \theta\)[/tex], [tex]\(\cot \theta\)[/tex], and find out in which quadrant [tex]\(\theta\)[/tex] lies given that [tex]\(\csc \theta = -\frac{5}{4}\)[/tex] and [tex]\(\sec \theta > 0\)[/tex].
1. Finding [tex]\(\sin \theta\)[/tex]:
- Recall that [tex]\(\csc \theta = \frac{1}{\sin \theta}\)[/tex].
- We are given that [tex]\(\csc \theta = -\frac{5}{4}\)[/tex].
- Therefore, [tex]\(\sin \theta = \frac{1}{\csc \theta} = \frac{1}{-\frac{5}{4}} = -\frac{4}{5}\)[/tex].
- So, [tex]\(\sin \theta = -0.8\)[/tex].
2. Determining the Quadrant:
- Given that [tex]\(\sec \theta > 0\)[/tex], and since [tex]\(\sec \theta = \frac{1}{\cos \theta}\)[/tex], we know that [tex]\(\cos \theta > 0\)[/tex].
- [tex]\(\cos \theta\)[/tex] is positive in Quadrants I and IV.
- Since [tex]\(\sin \theta = -0.8\)[/tex] indicates a negative sine value, [tex]\(\theta\)[/tex] cannot be in Quadrant I.
- Therefore, [tex]\(\theta\)[/tex] must be in Quadrant IV.
3. Finding [tex]\(\cos \theta\)[/tex]:
- Use the Pythagorean identity: [tex]\(\sin^2 \theta + \cos^2 \theta = 1\)[/tex].
- Substitute [tex]\(\sin \theta = -0.8\)[/tex] into the identity:
[tex]\[ (-0.8)^2 + \cos^2 \theta = 1 \][/tex]
[tex]\[ 0.64 + \cos^2 \theta = 1 \][/tex]
[tex]\[ \cos^2 \theta = 1 - 0.64 \][/tex]
[tex]\[ \cos^2 \theta = 0.36 \][/tex]
- Take the positive square root, since [tex]\(\cos \theta\)[/tex] is positive in Quadrant IV:
[tex]\[ \cos \theta = \sqrt{0.36} = 0.6 \][/tex]
4. Finding [tex]\(\cot \theta\)[/tex]:
- Recall that [tex]\(\cot \theta = \frac{\cos \theta}{\sin \theta}\)[/tex].
- Substitute [tex]\(\cos \theta = 0.6\)[/tex] and [tex]\(\sin \theta = -0.8\)[/tex]:
[tex]\[ \cot \theta = \frac{0.6}{-0.8} = -\frac{3}{4} = -0.75 \][/tex]
So, the values are:
[tex]\[
\sin \theta = -0.8
\][/tex]
[tex]\[
\cot \theta = -0.75
\][/tex]
Final Answer:
[tex]\[
\sin \theta = -0.8, \quad \cot \theta = -0.75
\][/tex]
