Solve Equations with Inverse Tangent

The question asks for the angle whose tangent is 1.

Since tan(π/4) = 1 and π/4 lies within the range (−π/2, π/2),

arctan(1) = π/4.

We know tan(π/3) = √3.

For a negative value, reflect across the x-axis → tan(−π/3) = −√3.

Since −π/3 is within the range (−π/2, π/2), x = −π/3.

The arctangent function is restricted so it returns one unique angle.

That range is (−π/2, π/2), because tangent is undefined at ±π/2.

Hence, the correct range is (−π/2, π/2).

The arctan range is (−π/2, π/2), and the endpoints are excluded.

Thus, 0, −π/3, and π/6 are valid, but π/2 is not in the range.

tan(60°) = √3.

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

We want the angle whose tangent is 0.75.

Since tan(36°) is slightly less than 0.75 and tan(37°) is slightly greater than 0.75, the correct angle is close to 37°.

Rounded to the nearest tenth of a degree, arctan(0.75) ≈ 36.9°.

tan and arctan are inverse functions.

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

Hence, tan(arctan(−2)) = −2.

Start with the equation: 3tan(x) − √3 = 0

First, add √3 to both sides:

3tan(x) = √3

Then, divide both sides by 3:

tan(x) = √3 / 3

Then, we use the special-angle fact:

tan(π/6) = √3 / 3

So the principal-value angle satisfying tan(x) = √3/3 is:

x = π/6

This value lies in the arctan principal range (−π/2 to π/2).

tan(0) = 0.

So the angle whose tangent equals 0 is 0.

Therefore, arctan(0) = 0.

Use the tangent ratio: tan(θ) = opposite / adjacent

First, substitute the values:

tan(θ) = 7 / 7

tan(θ) = 1

Then, to express θ, take the inverse tangent:

θ = arctan(1)

Since arctan(1) = 45°, this is the correct expression for θ.

tan(−π/4) = −1.

Since −π/4 is within (−π/2, π/2), arctan(−1) = −π/4.

tan(x) = 0 when x = 0, π, 2π, etc.

The principal value (in −π/2 to π/2) is 0.

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

Thus, x = arctan(5).

We want the angle whose tangent is 4/3 ≈ 1.33.

tan(0.79 rad) ≈ 1

tan(1.05 rad) ≈ 1.73

Since 1.33 lies between 1 and 1.73, the angle must be between 0.79 and 1.05 radians.

The closest value is about 0.93 radians.

tan(arctan(x)) = x is always true because tan(arctan(x)) simply returns the original input x.

Whereas arctan(tan(x)) = x is not always true because arctan(tan(x)) only equals x when x is in the principal range (−π/2, π/2). Also, arctan(x) + arctan(1/x) = π/2 for all x is not always true; it holds only when x > 0.

and lastly arctan(x) = arcsin(x) is also false because arctan(x) and arcsin(x) are different functions with different domains and ranges.

Simplify: 2tan(θ) = 2√3 → tan(θ) = √3.

The angle whose tangent equals √3 is π/3.

Hence, θ = π/3.

tan(π/6) = √3/3.

Since the value is negative, the angle is −π/6 in the range (−π/2, π/2).

Thus, arctan(−√3/3) = −π/6.

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

Since tan(π/4) = 1, a = 1.

tan(θ) = rise/run = 18/36 = 1/2.

So θ = arctan(18/36).

This gives the correct angle of elevation.

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 within the principal range of arctan, which is between −90° and 90°.