Test the polar equation for symmetry with respect to the polar axis, the pole, and the line $\theta = \pi / 2$ . $r = 3 \cos \theta \csc \theta$
I $\text { symmetric about the polar axis }$
II $\text { symmetric about the pole }$
III $\text { symmetric about the line } \theta = \pi / 2$
A)I only
B)I and II
C) I and III
D)II and III
E) none of these

Question

Quizplus · Accepted Answer

To test for symmetry in polar coordinates:- Symmetry about the polar axis (x-axis) is tested by replacing $\theta$ with $-\theta$. If the equation remains unchanged, it is symmetric about the polar axis.- Symmetry about the pole is tested by replacing $r$ with $-r$. If the equation remains unchanged, it is symmetric about the pole.- Symmetry about the line $\theta = \pi / 2$ is tested by replacing $\theta$ with $\pi - \theta$. If the equation remains unchanged, it is symmetric about the line $\theta = \pi / 2$.Given $r = 3 \cos \theta \csc \theta$, let's test each condition:1. Replace $\theta$ with $-\theta$: $r = 3 \cos(-\theta) \csc(-\theta)$. Since $\cos(-\theta) = \cos(\theta)$ and $\csc(-\theta) = -\csc(\theta)$, the equation becomes $r = -3 \cos \theta \csc \theta$, which is not the same as the original equation. So, it is not symmetric about the polar axis.2. Replace $r$ with $-r$: $-r = 3 \cos \theta \csc \theta$, which does not revert to the original equation. So, it is not symmetric about the pole.3. Replace $\theta$ with $\pi - \theta$: $r = 3 \cos(\pi - \theta) \csc(\pi - \theta)$. Since $\cos(\pi - \theta) = -\cos(\theta)$ and $\csc(\pi - \theta) = \csc(\theta)$, the equation becomes $r = -3 \cos \theta \csc \theta$, which is not the same as the original equation. So, it is not symmetric about the line $\theta = \pi / 2$.Therefore, the equation does not exhibit any of the tested symmetries.

Test the Polar Equation for Symmetry with Respect to the Polar