Inverse Tangent: Evaluate & Interpret (Principal Values)

  • 9th Grade
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| Questions: 20 | Updated: Dec 11, 2025
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1) Evaluate arctan(1) in radians.

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

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

Explanation

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)

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

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.

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5) Evaluate arctan(√3) in degrees.

Explanation

tan(60°) = √3.

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

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

Explanation

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°

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7) Evaluate tan(arctan(−2)).

Explanation

The tangent and arctangent functions are inverses.

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

Thus, tan(arctan(−2)) = −2.

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

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.

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9) Evaluate arctan(0) in radians.

Explanation

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

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

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

Explanation

tan(θ) = opposite/adjacent = 7/7 = 1.

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

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11) Evaluate arctan(−1) in radians.

Explanation

tan(−π/4) = −1.

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

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12) Find the principal value solving tan(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.

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

Explanation

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

Therefore, x = arctan(5).

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

Explanation

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

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

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.

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

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

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.

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18) If arctan(a) = π/4, find a.

Explanation

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

Since tan(π/4) = 1, a = 1.

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

Explanation

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)

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20) Evaluate arctan(√3/3) in degrees.

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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Evaluate arctan(1) in radians.
Solve for x in the principal range: tan(x) = −√3.
What is the range of y = arctan(x)?
Which value is NOT in the range of y = arctan(x)?
Evaluate arctan(√3) in degrees.
If tan(θ) = 0.75 and θ = arctan(0.75), what is θ to...
Evaluate tan(arctan(−2)).
Solve 3tan(x) − √3 = 0 for the principal value of x.
Evaluate arctan(0) in radians.
A right triangle has opposite side 7 and adjacent side 7 relative to...
Evaluate arctan(−1) in radians.
Find the principal value solving tan(x) = 0.
Find the principal value of x given tan(x) = 5.
Evaluate arctan(4/3) to the nearest 0.01 radians.
Which identity is always true for all real x?
Solve for θ in (−π/2, π/2): 2tan(θ) = 2√3.
Evaluate arctan(−√3/3) in radians.
If arctan(a) = π/4, find a.
A ramp rises 18 inches over a horizontal run of 36 inches. Let θ...
Evaluate arctan(√3/3) in degrees.
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