Inverse Tangent: Evaluate & Interpret (Principal Values)

9th Grade

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Quizzes Created: 8156 | Total Attempts: 9,588,805

| Attempts: 19 | Questions: 20 | Updated: Jan 22, 2026

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1) A right triangle has opposite side 7 and adjacent side 7 relative to angle θ. Express θ using inverse tangent.

Arctan(0)

Arctan(1)

Arctan(7)

Arctan(14)

tan(θ) = opposite/adjacent = 7/7 = 1.

Therefore, θ = arctan(1), which equals π/4 radians or 45°.

Explanation

tan(θ) = opposite/adjacent = 7/7 = 1.

Therefore, θ = arctan(1), which equals π/4 radians or 45°.

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About This Quiz

Inverse Tangent: Evaluate & Interpret (Principal Values) - Quiz

Ready to explore how the arctangent (arctan) function links slopes, ratios, and angles? This quiz focuses on evaluating arctan values, identifying the correct principal range (–π/2 to π/2), and interpreting what each result represents on the unit circle. You will practice determining when tangent is positive or negative, finding corresponding... see moreangles in radians or degrees, and applying tan(arctan x) = x to verify results. By the end, you will build a strong conceptual understanding of inverse tangent relationships.
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2) Evaluate arctan(1) in radians.

π/6

π/4

π/3

π/2

arctan(1) asks for the angle whose tangent is 1.

On the unit circle, tan(π/4) = 1.

Because π/4 lies in the principal range (−π/2, π/2), arctan(1) = π/4.

Explanation

arctan(1) asks for the angle whose tangent is 1.

On the unit circle, tan(π/4) = 1.

Because π/4 lies in the principal range (−π/2, π/2), arctan(1) = π/4.

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3) Solve for x in the principal range: tan(x) = −√3.

−π/6

−π/4

−π/3

−π/2

We know tan(π/3) = √3.

Since tangent is an odd function (tan(−θ) = −tan(θ)), tan(−π/3) = −√3.

Thus, in the principal range (−π/2, π/2), x = −π/3.

Explanation

We know tan(π/3) = √3.

Since tangent is an odd function (tan(−θ) = −tan(θ)), tan(−π/3) = −√3.

Thus, in the principal range (−π/2, π/2), x = −π/3.

Submit

4) What is the range of y = arctan(x)?

[0, π]

[−π/2, π/2]

(−π/2, π/2)

(−∞, ∞)

The arctan (inverse tangent) function returns the angle whose tangent equals x.

Tangent can take any real value, but the angle that arctan produces is always restricted to:

greater than −π/2

less than π/2

Arctan never actually reaches ±π/2; it only approaches them.

So the range is:

y ∈ (−π/2, π/2)

Explanation

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5) Which value is NOT in the range of y = arctan(x)?

π/2

−π/3

π/6

arctan(x) only gives angles between −π/2 and π/2 (not including those endpoints).

0, −π/3, and π/6 are within that range, but π/2 is not.

Explanation

arctan(x) only gives angles between −π/2 and π/2 (not including those endpoints).

0, −π/3, and π/6 are within that range, but π/2 is not.

Submit

6) Evaluate arctan(√3) in degrees.

60°

45°

30°

90°

tan(60°) = √3.

Because 60° lies in the arctan range (−90°, 90°), arctan(√3) = 60°.

Explanation

tan(60°) = √3.

Because 60° lies in the arctan range (−90°, 90°), arctan(√3) = 60°.

Submit

7) If tan(θ) = 0.75 and θ = arctan(0.75), what is θ to the nearest tenth of a degree?

36.9°

45.0°

53.1°

60.0°

We are given: tan(θ) = 0.75

θ = arctan(0.75)

To estimate the angle, note the following benchmark values:

tan(36°) ≈ 0.73

tan(37°) ≈ 0.75

tan(38°) ≈ 0.78

Since 0.75 is closest to tan(37°), the angle must be very close to 37°.

So, rounding to the nearest tenth:

θ ≈ 36.9°

Explanation

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8) Evaluate tan(arctan(−2)).

−2

1/2

The tangent and arctangent functions are inverses.

So tan(arctan(x)) = x for all real x.

Thus, tan(arctan(−2)) = −2.

Explanation

The tangent and arctangent functions are inverses.

So tan(arctan(x)) = x for all real x.

Thus, tan(arctan(−2)) = −2.

Submit

9) Solve 3tan(x) − √3 = 0 for the principal value of x.

X = π/2

X = π/4

X = π/3

X = π/6

Simplify: 3tan(x) − √3 = 0 → tan(x) = √3/3 = 1/√3.

The angle whose tangent is 1/√3 is π/6.

Thus, x = π/6.

Explanation

Simplify: 3tan(x) − √3 = 0 → tan(x) = √3/3 = 1/√3.

The angle whose tangent is 1/√3 is π/6.

Thus, x = π/6.

Submit

10) Evaluate arctan(0) in radians.

π/4

π/2

tan(0) = 0, so arctan(0) = 0.

This is the center of the principal range (−π/2, π/2).

Explanation

tan(0) = 0, so arctan(0) = 0.

This is the center of the principal range (−π/2, π/2).

Submit

11) Evaluate arctan(−1) in radians.

−π/6

−π/4

π/4

tan(−π/4) = −1.

Since −π/4 lies within the arctan range (−π/2, π/2), arctan(−1) = −π/4.

Explanation

tan(−π/4) = −1.

Since −π/4 lies within the arctan range (−π/2, π/2), arctan(−1) = −π/4.

Submit

12) Find the principal value solving tan(x) = 0.

π/2

−π/2

tan(x) = 0 when x = 0, π, 2π, etc.

The principal value (the one in −π/2 to π/2) is x = 0.

Explanation

tan(x) = 0 when x = 0, π, 2π, etc.

The principal value (the one in −π/2 to π/2) is x = 0.

Submit

13) Find the principal value of x given tan(x) = 5.

X = −arctan(5)

X = arctan(5)

X = π − arctan(5)

No solution

Arctangent gives the unique angle between −π/2 and π/2 whose tangent is 5.

Therefore, x = arctan(5).

Explanation

Arctangent gives the unique angle between −π/2 and π/2 whose tangent is 5.

Therefore, x = arctan(5).

Submit

14) Evaluate arctan(4/3) to the nearest 0.01 radians.

0.93

0.64

0.73

1.33

We want the angle whose tangent is 4/3.

tan(0.79 rad) ≈ 1

tan(1.05 rad) ≈ 1.73

Since 4/3 ≈ 1.33 is between 1 and 1.73, the angle must be between 0.79 and 1.05 radians.

The value that matches tan(x) = 4/3 is approximately:

x ≈ 0.93 radians

Explanation

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15) Which identity is always true for all real x?

Arctan(tan(x)) = x

Arctan(x) + arctan(1/x) = π/2 for all x

Arctan(x) = arcsin(x)

Tan(arctan(x)) = x

tan and arctan undo each other for any real x.

The identity tan(arctan(x)) = x always holds true.

The others only hold in limited ranges or are incorrect.

Explanation

tan and arctan undo each other for any real x.

The identity tan(arctan(x)) = x always holds true.

The others only hold in limited ranges or are incorrect.

Submit

16) Solve for θ in (−π/2, π/2): 2tan(θ) = 2√3.

θ = π/6

θ = π/4

θ = π/3

θ = arctan(2√3)

Divide both sides by 2 → tan(θ) = √3.

The angle whose tangent is √3 in the principal range is π/3.

Therefore, θ = π/3.

Explanation

Divide both sides by 2 → tan(θ) = √3.

The angle whose tangent is √3 in the principal range is π/3.

Therefore, θ = π/3.

Submit

17) Evaluate arctan(−√3/3) in radians.

−π/6

−π/4

−π/3

−π/2

tan(π/6) = √3/3.

For a negative value, take the reflection below the x-axis: θ = −π/6.

Thus, arctan(−√3/3) = −π/6.

Explanation

tan(π/6) = √3/3.

For a negative value, take the reflection below the x-axis: θ = −π/6.

Thus, arctan(−√3/3) = −π/6.

Submit

18) If arctan(a) = π/4, find a.

1/2

√3/2

√2/2

Apply tangent to both sides: a = tan(π/4).

Since tan(π/4) = 1, a = 1.

Explanation

Apply tangent to both sides: a = tan(π/4).

Since tan(π/4) = 1, a = 1.

Submit

19) A ramp rises 18 inches over a horizontal run of 36 inches. Let θ be the angle of elevation. Which expression gives θ?

Arctan(36/18)

Arctan(2)

Arctan(1/3) + π

Arctan(18/36)

The ramp forms a right triangle where:

rise (vertical) = 18 inches

run (horizontal) = 36 inches

θ = angle of elevation

For ramps, the angle is found using tangent:

tan(θ) = opposite / adjacent

tan(θ) = 18 / 36

So the angle is:

θ = arctan(18/36)

Explanation

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20) Evaluate arctan(√3/3) in degrees.

15°

30°

45°

60°

We use a known special-angle fact: tan(30°) = √3/3.

So the angle whose tangent equals √3/3 is 30°.

This value is also in the principal range of arctan, which is between –90° and 90°.

Explanation

We use a known special-angle fact: tan(30°) = √3/3.

So the angle whose tangent equals √3/3 is 30°.

This value is also in the principal range of arctan, which is between –90° and 90°.

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