To determine which functions have a vertical asymptote for values of \( \theta \) such that \( \cos \theta = 1 \), we need to consider what happens at those values.
The values of \( \theta \) where \( \cos \theta = 1 \) occur at: \[ \theta = 2n\pi, \quad n \in \mathbb{Z} \] This corresponds to the angles where the cosine function reaches its maximum value of 1.
Now let's analyze each function for vertical asymptotes:
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\( y = \sin \theta \): The sine function does not have vertical asymptotes. It oscillates between -1 and 1 for all \( \theta \). So, no vertical asymptote here.
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\( y = \cos \theta \): As with the sine function, the cosine function also does not have vertical asymptotes and oscillates between -1 and 1. So, no vertical asymptote here.
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\( y = \tan \theta \): The tangent function has vertical asymptotes where \( \cos \theta = 0 \), not where \( \cos \theta = 1 \). So, no vertical asymptote here.
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\( y = \sec \theta \): The secant function is defined as \( \sec \theta = \frac{1}{\cos \theta} \). When \( \cos \theta = 1 \) (e.g., at \( \theta = 0, 2\pi, \ldots \)), the secant is defined and equals 1, so there are no vertical asymptotes for this function.
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\( y = \csc \theta \): The cosecant function is defined as \( \csc \theta = \frac{1}{\sin \theta} \). While this function has vertical asymptotes at \( \theta = n\pi \) (where \( \sin \theta = 0 \)), it does not have vertical asymptotes at points where \( \cos \theta = 1 \), so no vertical asymptote here.
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\( y = \cot \theta \): The cotangent function is \( \cot \theta = \frac{\cos \theta}{\sin \theta} \). It has vertical asymptotes where \( \sin \theta = 0 \). Since these asymptotes do not occur at points where \( \cos \theta = 1 \), there are no vertical asymptotes for this function either.
In conclusion, none of the provided functions have vertical asymptotes for values of \( \theta \) where \( \cos \theta = 1 \).
Answer: None of the functions have vertical asymptotes where \( \cos \theta = 1 \).
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