Final answer:
The polar equation R = 3 sec θ is symmetric with respect to the polar axis and the pole, but not with respect to the line θ = π/2.
Step-by-step explanation:
A polar equation is symmetric with respect to the polar axis if replacing θ with -θ does not change the equation. In this case, the equation is R = 3 sec θ. Replacing θ with -θ gives us R = 3 sec (-θ). Since the secant function is even, meaning sec (-θ) = sec θ, the equation remains the same. Therefore, the equation is symmetric with respect to the polar axis.
A polar equation is symmetric with respect to the pole if replacing R with -R does not change the equation. In this case, the equation is R = 3 sec θ. Replacing R with -R gives us -R = 3 sec θ. Since the secant function is an even function, the equation can be rewritten as R = -3 sec θ, which is equivalent to the original equation. Therefore, the equation is symmetric with respect to the pole.
A polar equation is symmetric with respect to the line θ = π/2 if replacing θ with π - θ does not change the equation. In this case, the equation is R = 3 sec θ. Replacing θ with π - θ gives us R = 3 sec (π - θ). Using the identity sec (π - θ) = -sec θ, we can rewrite the equation as R = -3 sec θ, which is not equivalent to the original equation. Therefore, the equation is not symmetric with respect to the line θ = π/2.