Given that [tex]\(\csc \theta = -\frac{5}{2}\)[/tex] and [tex]\(\theta\)[/tex] is in the third quadrant, let's find the remaining trigonometric functions ([tex]\(\sin \theta\)[/tex], [tex]\(\cos \theta\)[/tex], [tex]\(\tan \theta\)[/tex], [tex]\(\sec \theta\)[/tex], [tex]\(\cot \theta\)[/tex]) step by step.
### Step 1: Find [tex]\(\sin \theta\)[/tex]
The cosecant function is the reciprocal of the sine function, i.e.,
[tex]\[
\csc \theta = \frac{1}{\sin \theta}
\][/tex]
Given [tex]\(\csc \theta = -\frac{5}{2}\)[/tex], we can find [tex]\(\sin \theta\)[/tex] by taking the reciprocal:
[tex]\[
\sin \theta = \frac{1}{\csc \theta} = \frac{1}{-\frac{5}{2}} = -\frac{2}{5}
\][/tex]
### Step 2: Determine the sign of [tex]\(\sin \theta\)[/tex]
In the third quadrant, both sine and cosine functions are negative. Hence, [tex]\(\sin \theta\)[/tex] remains as [tex]\(-\frac{2}{5}\)[/tex].
### Step 3: Find [tex]\(\cos \theta\)[/tex]
Use the Pythagorean identity:
[tex]\[
\sin^2 \theta + \cos^2 \theta = 1
\][/tex]
Substitute [tex]\(\sin \theta\)[/tex]:
[tex]\[
\left( -\frac{2}{5} \right)^2 + \cos^2 \theta = 1
\][/tex]
[tex]\[
\frac{4}{25} + \cos^2 \theta = 1
\][/tex]
Solving for [tex]\(\cos^2 \theta\)[/tex]:
[tex]\[
\cos^2 \theta = 1 - \frac{4}{25} = \frac{25}{25} - \frac{4}{25} = \frac{21}{25}
\][/tex]
Taking the square root and considering the sign in the third quadrant:
[tex]\[
\cos \theta = -\sqrt{\frac{21}{25}} = -\frac{\sqrt{21}}{5}
\][/tex]
### Step 4: Find [tex]\(\tan \theta\)[/tex]
The tangent function is the ratio of the sine function to the cosine function:
[tex]\[
\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{-\frac{2}{5}}{-\frac{\sqrt{21}}{5}} = \frac{2}{\sqrt{21}} = \frac{2\sqrt{21}}{21}
\][/tex]
### Step 5: Find [tex]\(\sec \theta\)[/tex]
The secant function is the reciprocal of the cosine function:
[tex]\[
\sec \theta = \frac{1}{\cos \theta} = \frac{1}{-\frac{\sqrt{21}}{5}} = -\frac{5}{\sqrt{21}} = -\frac{5\sqrt{21}}{21}
\][/tex]
### Step 6: Find [tex]\(\cot \theta\)[/tex]
The cotangent function is the reciprocal of the tangent function:
[tex]\[
\cot \theta = \frac{1}{\tan \theta} = \frac{1}{\frac{2\sqrt{21}}{21}} = \frac{21}{2\sqrt{21}} = \frac{21}{2\sqrt{21}} \cdot \frac{\sqrt{21}}{\sqrt{21}} = \frac{21\sqrt{21}}{42} = \frac{\sqrt{21}}{2}
\][/tex]
So, we have:
[tex]\[
\sin \theta = -0.4, \quad \cos \theta = -0.916515138991168, \quad \tan \theta = 0.4364357804719848, \quad \sec \theta = -1.0910894511799618, \quad \cot \theta = 2.29128784747792.
\][/tex]
