Let n represent any integer.
tan x has a vertical asymptote at x = .

cot x has a vertical asymptote at x = .

sec x has a vertical asymptote at x = .

csc x has a vertical asymptote at x = .

Hello mate!

The asymptotes are located at the points where the denominator is equal to zero, then, let's solve all the cases:

A) We know that tan(x) = sen(x)/cos(x), so the asymptote is at the value of x that makes the cosine equal to zero.

This is pi/2

we know that tan(x) has an asymptote at pi/2, but because of the bheaviour of the trigonometric function, tan(x) also has an asimptote at pi/2 + pi.

then if we define n as an integer number, the asimpotes of tan(x) are when x = pi/2 + n*pi

B) now we ave cot(x) = 1/tan(x) = cos(x)/sin(x), so the asymptote is at the points where sin(x) is equal to zero, that are the points of the form x = pi + n*pi

C) Sec(x) = 1/cos(x), then the asympotes are, like in part A, at the values of x where cos(x) = 0, so the asympotetes are at the values of x = pi/2 + n*pi

D) cosec(x) = 1/sin(x), again, the asymptotes are when sin(x) = 0, so the asympotes are in the values of x = pi + n*pi

Where in all the situations n is an integer.

there is no graph so i cant sorry

step-by-step explanation:

step-by-step explanation:

your answer is a