Answer :
The question is incomplete. Here is the complete question:
Which of the functions have a range of all real numbers greater than or equal to 1 or less than or equal to -1? check all that apply.
A. [tex]y=\sec x[/tex]
B. [tex]y= \tan x[/tex]
C. [tex]y= \cot x[/tex]
D. [tex]y= \csc x[/tex]
Answer:
A. [tex]y=\sec x[/tex]
D. [tex]y=\csc x[/tex]
Step-by-step explanation:
Given:
The range is greater than or equal to 1 or less than or equal to -1.
The given choices are:
Choice A: [tex]y=\sec x[/tex]
We know that, the [tex]\sec x=\frac{1}{\cos x}[/tex]
The range of [tex]\cos x[/tex] is from -1 to 1 given as [-1, 1]. So,
[tex]|\cos x|\leq 1\\\textrm{Taking reciprocal, the inequality sign changes}\\\frac{1}{|\cos x|}\geq 1\\|\sec x|\geq 1[/tex]
Therefore, on removing the absolute sign, we rewrite the secant function as:
[tex]\sec x\leq -1\ or\ \sec x\geq 1\\[/tex]
Therefore, the range of [tex]y=\sec x[/tex] is all real numbers greater than or equal to 1 or less than or equal to-1.
Choice B: [tex]y= \tan x[/tex]
We know that, the range of tangent function is all real numbers. So, choice B is incorrect.
Choice C: [tex]y= \cot x[/tex]
We know that, the range of cotangent function is all real numbers. So, choice C is incorrect.
Choice D: [tex]y=\csc x[/tex]
We know that, the [tex]\csc x=\frac{1}{\sin x}[/tex]
The range of [tex]\sin x[/tex] is from -1 to 1 given as [-1, 1]. So,
[tex]|\sin x|\leq 1\\\textrm{Taking reciprocal, the inequality sign changes}\\\frac{1}{|\sin x|}\geq 1\\|\csc x|\geq 1[/tex]
Therefore, on removing the absolute sign, we rewrite the cosecant function as:
[tex]\csc x\leq -1\ or\ \csc x\geq 1\\[/tex]
Therefore, the range of [tex]y=\csc x[/tex] is all real numbers greater than or equal to 1 or less than or equal to-1.