Answer:
[tex]\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = \frac{5 - 3\sqrt 5}{5}[/tex]
Step-by-step explanation:
Given
[tex]\cos(\theta) = -\frac{2}{3}[/tex]
[tex]\theta \to[/tex] Quadrant III
Required
Determine [tex]\tan(\theta) \cdot \cot(\theta) + \csc(\theta)[/tex]
We have:
[tex]\cos(\theta) = -\frac{2}{3}[/tex]
We know that:
[tex]\sin^2(\theta) + \cos^2(\theta) = 1[/tex]
This gives:
[tex]\sin^2(\theta) + (-\frac{2}{3})^2 = 1[/tex]
[tex]\sin^2(\theta) + (\frac{4}{9}) = 1[/tex]
Collect like terms
[tex]\sin^2(\theta) = 1 - \frac{4}{9}[/tex]
Take LCM and solve
[tex]\sin^2(\theta) = \frac{9 -4}{9}[/tex]
[tex]\sin^2(\theta) = \frac{5}{9}[/tex]
Take the square roots of both sides
[tex]\sin(\theta) = \±\frac{\sqrt 5}{3}[/tex]
Sin is negative in quadrant III. So:
[tex]\sin(\theta) = -\frac{\sqrt 5}{3}[/tex]
Calculate [tex]\csc(\theta)[/tex]
[tex]\csc(\theta) = \frac{1}{\sin(\theta)}[/tex]
We have: [tex]\sin(\theta) = -\frac{\sqrt 5}{3}[/tex]
So:
[tex]\csc(\theta) = \frac{1}{-\frac{\sqrt 5}{3}}[/tex]
[tex]\csc(\theta) = \frac{-3}{\sqrt 5}[/tex]
Rationalize
[tex]\csc(\theta) = \frac{-3}{\sqrt 5}*\frac{\sqrt 5}{\sqrt 5}[/tex]
[tex]\csc(\theta) = \frac{-3\sqrt 5}{5}[/tex]
So, we have:
[tex]\tan(\theta) \cdot \cot(\theta) + \csc(\theta)[/tex]
[tex]\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = \tan(\theta) \cdot \frac{1}{\tan(\theta)} + \csc(\theta)[/tex]
[tex]\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = 1 + \csc(\theta)[/tex]
Substitute: [tex]\csc(\theta) = \frac{-3\sqrt 5}{5}[/tex]
[tex]\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = 1 -\frac{3\sqrt 5}{5}[/tex]
Take LCM
[tex]\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = \frac{5 - 3\sqrt 5}{5}[/tex]
