Question

question 2 (1 point)
which trigonometric function has zeros at ( x = npi,nin z ), and has vertical asymptotes midway between the zeros?
( \bigcirc ) a) ( y=sec x )
( \bigcirc ) b) ( y=cot x )
( \bigcirc ) c) ( y=csc x )
( \bigcirc ) d) ( y=\tan x )

Explanation:

• For \(y = \sec x=\frac{1}{\cos x}\), the zeros of \(\cos x\) are \(x=(2n + 1)\frac{\pi}{2},n\in Z\), and \(\sec x\) has vertical asymptotes at \(x = n\pi,n\in Z\).
• For \(y=\cot x=\frac{\cos x}{\sin x}\), the zeros of \(\cot x\) occur when \(\cos x = 0\), i.e., \(x=(2n + 1)\frac{\pi}{2},n\in Z\), and vertical asymptotes when \(\sin x=0\), i.e., \(x = n\pi,n\in Z\).
• For \(y=\csc x=\frac{1}{\sin x}\), the zeros of \(\sin x\) are \(x = n\pi,n\in Z\), and \(\csc x\) has vertical asymptotes at \(x=(2n + 1)\frac{\pi}{2},n\in Z\).
• For \(y = \tan x=\frac{\sin x}{\cos x}\), the zeros of \(\tan x\) occur when \(\sin x=0\), i.e., \(x = n\pi,n\in Z\), and vertical asymptotes when \(\cos x = 0\), i.e., \(x=(2n + 1)\frac{\pi}{2},n\in Z\) (mid - way between \(x = n\pi\) and \(x=(n + 1)\pi\)).

Answer:

d) \(y=\tan x\)