Question

question 6 (1 point)
the graph shown is of the function
a) ( y = \frac { 1 } { cos x })
b) ( y = \frac { 1 } { \tan x })
c) ( y = \frac { 1 } { sec x })
d) ( y = \frac { 1 } { sin x })

Explanation:

Step1: Recall trigonometric identities

We know that \(\sec x=\frac{1}{\cos x}\), \(\cot x = \frac{1}{\tan x}\), \(\cos x=\frac{1}{\sec x}\), \(\csc x=\frac{1}{\sin x}\).

Step2: Analyze the vertical asymptotes

The vertical asymptotes of \(y = \cos x\) occur at \(x=(2n + 1)\frac{\pi}{2},n\in\mathbb{Z}\). For \(y=\frac{1}{\cos x}\), the vertical asymptotes are also at \(x=(2n + 1)\frac{\pi}{2},n\in\mathbb{Z}\).

• For \(y = \tan x\), vertical asymptotes at \(x=(2n + 1)\frac{\pi}{2},n\in\mathbb{Z}\), but \(y=\frac{1}{\tan x}=\cot x\) has vertical asymptotes at \(x = n\pi,n\in\mathbb{Z}\).
• For \(y=\sec x=\frac{1}{\cos x}\), \(y = \cos x\) has zeros at \(x=(2n + 1)\frac{\pi}{2},n\in\mathbb{Z}\), so \(y=\frac{1}{\cos x}\) has vertical asymptotes at \(x=(2n + 1)\frac{\pi}{2},n\in\mathbb{Z}\)
• For \(y=\sin x\), vertical asymptotes of \(y=\frac{1}{\sin x}\) (i.e., \(y = \csc x\)) occur at \(x=n\pi,n\in\mathbb{Z}\)

Step3: Analyze the value at \(x = 0\)

When \(x = 0\), \(y=\frac{1}{\cos(0)}=\frac{1}{1}=1\), \(y=\frac{1}{\tan(0)}\) is undefined (since \(\tan(0)=0\)), \(y=\frac{1}{\sec(0)}=\cos(0) = 1\) (but \(y=\frac{1}{\sec x}=\cos x\) is a cosine - type curve, not the curve in the graph), \(y=\frac{1}{\sin(0)}\) is undefined.

Answer:

A. \(y=\frac{1}{\cos x}\)