where are the vertical asymptotes on the graph of the secant function?\n$x = \\frac{\\pi}{2}+n\\pi$\n$x =…

where are the vertical asymptotes on the graph of the secant function?\n$x = \\frac{\\pi}{2}+n\\pi$\n$x = n\\pi$\n$x = 2n\\pi$\ndone

where are the vertical asymptotes on the graph of the secant function?\n$x = \\frac{\\pi}{2}+n\\pi$\n$x = n\\pi$\n$x = 2n\\pi$\ndone

Answer

Answer:

A. ( x=\frac{\pi}{2}+n\pi )

Explanation:

Step1: Recall the definition of the secant function

The secant function is ( y = \sec(x)=\frac{1}{\cos(x)} ).

Step2: Find when the denominator is zero

Vertical asymptotes occur when the denominator of a rational function is zero. We set ( \cos(x)=0 ).

Step3: Solve ( \cos(x) = 0 )

The solutions of ( \cos(x)=0 ) are ( x=\frac{\pi}{2}+n\pi ), where ( n\in\mathbb{Z} ). So the vertical asymptotes of ( y = \sec(x) ) are ( x=\frac{\pi}{2}+n\pi ).