The question is an illustration of trigonometry identity
The value of [tex]\theta[/tex] is [tex]337.38^o[/tex] and [tex]\cot(\theta) = -\frac{12}{5}[/tex]
The question is a not clear, so I will make use of [tex]\cos \theta = \frac{12}{13}[/tex].
Using:
[tex]\sin^2 \theta + \cos^2 \theta = 1[/tex]
Substitute [tex]\cos \theta = \frac{12}{13}[/tex]
[tex]\sin^2 \theta + (\frac{12}{13})^2 = 1[/tex]
Collect like terms
[tex]\sin^2 \theta = 1- (\frac{12}{13})^2[/tex]
[tex]\sin^2 \theta = 1- \frac{144}{169}[/tex]
Take LCM
[tex]\sin^2 \theta = \frac{169 - 144}{169}[/tex]
[tex]\sin^2 \theta = \frac{25}{169}[/tex]
Take square roots
[tex]\sin \theta = \±\frac{5}{13}[/tex]
From the question, we have cosine to be positive and [tex]\cot \theta < 0[/tex]
This means that [tex]\theta[/tex] is in the fourth quadrant, where [tex]\sin(\theta) < 0[/tex]
So, we have:
[tex]\sin \theta = -\frac{5}{13}[/tex]
The value of [tex]\cot(\theta)[/tex] is the
[tex]\cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)}[/tex]
Substitute [tex]\sin \theta = -\frac{5}{13}[/tex] and [tex]\cos \theta = \frac{12}{13}[/tex]
[tex]\cot(\theta) = \frac{12}{13} \div -\frac{5}{13}[/tex]
Rewrite as:
[tex]\cot(\theta) = \frac{12}{13} \times -\frac{13}{5}[/tex]
[tex]\cot(\theta) = -\frac{12}{5}[/tex]
The value of [tex]\theta[/tex] is calculated as follows:
[tex]\sin \theta = -\frac{5}{13}[/tex]
Take arcsin of both sides
[tex]\theta = sin^{-1}(-5/13)[/tex]
Because [tex]\theta[/tex] is in the 4th quadrant, the above equation becomes
[tex]\theta = 360^o + sin^{-1}(-5/13)[/tex]
[tex]\theta = 360^o - 22.62^o[/tex]
[tex]\theta = 337.38^o[/tex]
Hence, the value of [tex]\theta[/tex] is [tex]337.38^o[/tex] and [tex]\cot(\theta) = -\frac{12}{5}[/tex]
Read more about trigonometry at:
https://brainly.com/question/7565794
