Question

question 8 (1 point)
which trigonometric function has zeros at ( x = npi ), ( nin z ), and has vertical asymptotes midway between the zeros?
( \bigcirc ) a) ( y=cot x )
( \bigcirc ) b) ( y=\tan x )
( \bigcirc ) c) ( y=sec x )
( \bigcirc ) d) ( y=csc x )

Explanation:

• For \(y = \cot x=\frac{\cos x}{\sin x}\), the zeros occur when \(\cos x = 0\) (i.e., \(x=(n +\frac{1}{2})\pi,n\in Z\)) and vertical asymptotes when \(\sin x=0\) (i.e., \(x = n\pi,n\in Z\)).
• For \(y=\tan x=\frac{\sin x}{\cos x}\), the zeros occur when \(\sin x = 0\) (i.e., \(x = n\pi,n\in Z\)) and vertical asymptotes when \(\cos x=0\) (i.e., \(x=(n+\frac{1}{2})\pi,n\in Z\)).
• For \(y = \sec x=\frac{1}{\cos x}\), there are no zeros (since the numerator is \(1\)) and vertical asymptotes when \(\cos x = 0\) (i.e., \(x=(n+\frac{1}{2})\pi,n\in Z\)).
• For \(y=\csc x=\frac{1}{\sin x}\), there are no zeros (since the numerator is \(1\)) and vertical asymptotes when \(\sin x = 0\) (i.e., \(x = n\pi,n\in Z\)).

Answer:

B. \(y = \tan x\)