Secant Range Quiz: Secant Range & Excluded Values

1) Why are there “gaps” between the branches of the secθ graph around y ∈ (−1, 1)?

Because secθ cannot take values with |y| < 1

Because sinθ has period 2π

Because tanθ is undefined at π/2 + kπ

Because cosθ is unbounded

Since −1 ≤ cosθ ≤ 1, the reciprocal secθ satisfies |secθ| ≥ 1. Values between −1 and 1 are excluded, leaving horizontal gaps in the graph.

Explanation

Since −1 ≤ cosθ ≤ 1, the reciprocal secθ satisfies |secθ| ≥ 1. Values between −1 and 1 are excluded, leaving horizontal gaps in the graph.

2) If cosθ = 1/2, then secθ equals

2 (which is in the allowed range)

1/2 (which is in the allowed range)

2 (which is not allowed)

−2 (which is not allowed)

secθ = 1/cosθ = 1/(1/2) = 2. Since |secθ| ≥ 1, the value 2 is valid and lies in the range.

Explanation

secθ = 1/cosθ = 1/(1/2) = 2. Since |secθ| ≥ 1, the value 2 is valid and lies in the range.

3) Which of the following y-values are possible for y = secθ?

-2

−1/2

1

√2

-1

Possible values must satisfy y ≤ −1 or y ≥ 1. Thus −2, 1, √2, and −1 are valid; −1/2 is not.

Explanation

Possible values must satisfy y ≤ −1 or y ≥ 1. Thus −2, 1, √2, and −1 are valid; −1/2 is not.

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4) Does secθ = −1/2 have a real solution?

No, because |−1/2| < 1

Yes, when θ = 2π/3

Yes, when θ = π/3

Yes, when θ = π/2

Any solution would require |secθ| ≥ 1. Since |−1/2| < 1, there is no real θ satisfying secθ = −1/2.

Explanation

Any solution would require |secθ| ≥ 1. Since |−1/2| < 1, there is no real θ satisfying secθ = −1/2.

5) State the domain of y = secθ in set-builder form.

secθ is undefined where cosθ = 0, i.e., θ = π/2 + kπ. Excluding these points gives the domain.

Explanation

secθ is undefined where cosθ = 0, i.e., θ = π/2 + kπ. Excluding these points gives the domain.

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6) For y = 1/3, which equation has no real solution?

Secθ = y

Cosθ = y

Sinθ = y

Tanθ = y

cos and sin can take any value in [−1,1]; tan can take any real value. But secθ cannot equal 1/3 since |1/3| < 1.

Explanation

cos and sin can take any value in [−1,1]; tan can take any real value. But secθ cannot equal 1/3 since |1/3| < 1.

7) If cosθ = a with −1 ≤ a ≤ 1 and a ≠ 0, then secθ = ____ and |secθ| ≥ 1.

By definition, secθ = 1/cosθ = 1/a; since |a| ≤ 1 and a ≠ 0, we have |1/a| ≥ 1.

Explanation

By definition, secθ = 1/cosθ = 1/a; since |a| ≤ 1 and a ≠ 0, we have |1/a| ≥ 1.

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8) |secθ| ≥ 1 whenever secθ is defined.

True

False

Because |cosθ| ≤ 1 and secθ = 1/cosθ (with cosθ ≠ 0), reciprocation gives |secθ| ≥ 1.

Explanation

Because |cosθ| ≤ 1 and secθ = 1/cosθ (with cosθ ≠ 0), reciprocation gives |secθ| ≥ 1.

9) Select all equivalent ways to express the range condition for secθ.

|secθ| ≥ 1

Sec^2θ ≥ 1

Sec^2θ − 1 = tan^2θ ≥ 0 (where defined)

−1 < secθ < 1

Range is all real numbers

Since secθ = 1/cosθ, |secθ| ≥ 1. Squaring yields sec^2θ ≥ 1. Also sec^2θ − 1 = tan^2θ ≥ 0. The other two statements are false.

Explanation

Since secθ = 1/cosθ, |secθ| ≥ 1. Squaring yields sec^2θ ≥ 1. Also sec^2θ − 1 = tan^2θ ≥ 0. The other two statements are false.

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10) On an interval where cosθ > 0, what is the smallest value attained by secθ?

1 at θ = 2kπ

0 at θ = kπ

−1 at θ = (2k+1)π

It has no minimum

When cosθ > 0, secθ ≥ 1 with equality when cosθ = 1, occurring at θ = 2kπ (i.e., kπ where cos is +1).

Explanation

When cosθ > 0, secθ ≥ 1 with equality when cosθ = 1, occurring at θ = 2kπ (i.e., kπ where cos is +1).

11) Select all correct implications of the range for graphing y = secθ.

There are horizontal gaps for −1 < y < 1

Each branch stays entirely above y = 1 or entirely below y = −1

Vertices occur at y = ±1

The graph crosses y = 0 at θ = kπ

Vertical asymptotes occur where cosθ = 0

Because |secθ| ≥ 1, the graph leaves an empty band (−1,1). Branches have a unique extremum at y=±1 and are separated by asymptotes at cosθ=0.

Explanation

Because |secθ| ≥ 1, the graph leaves an empty band (−1,1). Branches have a unique extremum at y=±1 and are separated by asymptotes at cosθ=0.

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12) The equation secθ = 0 has no real solution.

True

False

secθ = 1/cosθ cannot be zero since 1 divided by any finite nonzero number is nonzero; and cosθ cannot be infinite.

Explanation

secθ = 1/cosθ cannot be zero since 1 divided by any finite nonzero number is nonzero; and cosθ cannot be infinite.

13) Select all true statements about secθ.

Secθ = 1/cosθ

Secθ is undefined where cosθ = 0

|secθ| < 1 at some θ

Range of secθ is y ≤ −1 or y ≥ 1

Secθ crosses y = 0 for some θ

Definitions and reciprocal reasoning: A and B hold. The range excludes (−1,1), so D is true. secθ never equals 0 and |secθ| is never < 1.

Explanation

Definitions and reciprocal reasoning: A and B hold. The range excludes (−1,1), so D is true. secθ never equals 0 and |secθ| is never < 1.

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14) What is the range of y = secθ (where defined)?

(−∞, −1] ∪ [1, ∞)

[−1, 1]

(−1, 1)

All real numbers

Since −1 ≤ cosθ ≤ 1 and secθ = 1/cosθ (when cosθ ≠ 0), taking reciprocals yields |secθ| ≥ 1. Thus y ≤ −1 or y ≥ 1.

Explanation

Since −1 ≤ cosθ ≤ 1 and secθ = 1/cosθ (when cosθ ≠ 0), taking reciprocals yields |secθ| ≥ 1. Thus y ≤ −1 or y ≥ 1.

15) Select all θ in [0, 2π) for which secθ ≥ 1.

θ = 0

θ = π/3

θ = π/2

θ = 2π/3

θ = 11π/6

Compute cos and reciprocals: cos0=1 ⇒ sec0=1; cos(π/3)=1/2 ⇒ sec=2; cos(11π/6)=√3/2 ⇒ sec≈1.155. At π/2, sec undefined; at 2π/3, cos negative so sec<−1.

Explanation

Compute cos and reciprocals: cos0=1 ⇒ sec0=1; cos(π/3)=1/2 ⇒ sec=2; cos(11π/6)=√3/2 ⇒ sec≈1.155. At π/2, sec undefined; at 2π/3, cos negative so sec<−1.

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