Question

which are vertical asymptotes for y = sec x?
π, π/4, 3π/4 π, π/6, 5π/6
-π/2, π/2, 3π/2 -π, π/3, 5π/3

Explanation:

Step1: Recall the definition of secant function

The secant function is defined as \( \sec x=\frac{1}{\cos x} \). Vertical asymptotes of a function occur where the function is undefined. For \( \sec x \), it is undefined when \( \cos x = 0 \).

Step2: Find the values where \( \cos x=0 \)

We know that \( \cos x = 0 \) when \( x=\frac{\pi}{2}+n\pi \), where \( n \) is an integer (..., -2, -1, 0, 1, 2,...).

Step3: Check the options

• For the first option: \( \pi,\frac{\pi}{4},\frac{3\pi}{4} \). \( \cos(\pi)= - 1

eq0 \), \( \cos(\frac{\pi}{4})=\frac{\sqrt{2}}{2}
eq0 \), \( \cos(\frac{3\pi}{4})=-\frac{\sqrt{2}}{2}
eq0 \), so no vertical asymptotes here.

• For the second option: \( \pi,\frac{\pi}{6},\frac{5\pi}{6} \). \( \cos(\pi)=-1

eq0 \), \( \cos(\frac{\pi}{6})=\frac{\sqrt{3}}{2}
eq0 \), \( \cos(\frac{5\pi}{6})=-\frac{\sqrt{3}}{2}
eq0 \), so no vertical asymptotes here.

• For the third option: \( -\frac{\pi}{2},\frac{\pi}{2},\frac{3\pi}{2} \).
• \( \cos(-\frac{\pi}{2}) = 0 \) (since \( \cos(-\theta)=\cos\theta \) and \( \cos(\frac{\pi}{2}) = 0 \))
• \( \cos(\frac{\pi}{2})=0 \)
• \( \cos(\frac{3\pi}{2}) = 0 \) (because \( \frac{3\pi}{2}=\frac{\pi}{2}+\pi \), and \( \cos(\frac{\pi}{2}+n\pi)=0 \) for integer \( n \))
• For the fourth option: \( -\pi,\frac{\pi}{3},\frac{5\pi}{3} \). \( \cos(-\pi)=-1

eq0 \), \( \cos(\frac{\pi}{3})=\frac{1}{2}
eq0 \), \( \cos(\frac{5\pi}{3})=\frac{1}{2}
eq0 \), so no vertical asymptotes here.

Answer:

\( -\frac{\pi}{2},\frac{\pi}{2},\frac{3\pi}{2} \) (the third option: \( -\frac{\pi}{2},\frac{\pi}{2},\frac{3\pi}{2} \))