We are given the values of [tex]\(\tan \theta = -\frac{12}{5}\)[/tex] and [tex]\(\csc \theta = -\frac{13}{5}\)[/tex], and the terminal point determined by [tex]\(\theta\)[/tex] is in the third quadrant. Let's analyze each of the statements given:
Statement A: "Cannot be true because if [tex]\(\tan \theta = -\frac{12}{5}\)[/tex], then [tex]\(\csc \theta = \pm \frac{13}{12}\)[/tex]."
Analysis:
- The tangent of an angle is given by the ratio [tex]\(\frac{\sin \theta}{\cos \theta}\)[/tex].
- The cosecant of an angle is [tex]\( \csc \theta = \frac{1}{\sin \theta} \)[/tex].
- For [tex]\(\csc \theta = -\frac{13}{5}\)[/tex], [tex]\(\sin \theta = -\frac{5}{13} \)[/tex].
Since [tex]\(\tan \theta\)[/tex] and [tex]\(\csc \theta\)[/tex] are given, if we had a corrected \csc \theta value as [tex]\(\pm \frac{13}{12}\)[/tex], it contradicts the given [tex]\(\sin \theta = -\frac{5}{13}\)[/tex]. Thus, it is not possible.
Statement A is False.
Statement B: "Cannot be true because [tex]\(\tan \theta\)[/tex] is greater than zero in quadrant 3."
Analysis:
- In the third quadrant, both sine and cosine are negative, hence [tex]\(\tan \theta = \frac{\sin \theta}{\cos \theta}\)[/tex] would be positive.
- Given [tex]\(\tan \theta = -\frac{12}{5}\)[/tex], which is negative, indicating that [tex]\(\tan \theta\)[/tex] cannot be positive.
This contradicts the statement that [tex]\(\tan \theta\)[/tex] must be positive in the third quadrant.
Statement B is True.
Statement C: "Cannot be true because [tex]\(\tan \theta\)[/tex] must be less than 1."
Analysis:
- [tex]\(\tan \theta = -\frac{12}{5}\)[/tex], which is significantly less than -1. This statement implies an impossibility.
Statement C is False.
Statement D: "Cannot be true because [tex]\(12^2 + 5^2 \neq 1\)[/tex]."
Analysis:
- If we consider the Pythagorean identity for the sine and cosine values forming the tangent value:
[tex]\[\sin^2 \theta + \cos^2 \theta = 1\][/tex]
The tangent formed combination does not form this relation.
Since the given [tex]\(\tan \theta\)[/tex] is using a different vale combination, this implies for the doubable scenario.
Statement D is True.
Thus, the solution yields:
- Statement A: False
- Statement B: True
- Statement C: False
- Statement D: True
