Evaluate the following expression. Your answer must be in exact form: for example, type pi/6 for π6 or DNE if the expression is undefined.

arctan(tan(−31π6))

Since arctan and tan are inverse operations, then logically

arctan(tan(-31π/6)) = -31π/6

however .... since it probably wants the angle closest to the origin, and
since tan(-31π/6) = tan(-π/6)

arctan(tan(-31π/6)
= arctan(tan(-π/6)
= -π/6

I am just wondering how you find tan(-π/6) from tan(-31π/6)

one rotation is 2π

so -31π/6 ÷ 2π = -5 - π/6
so we have 5 clockwise rotations and another clockwise of π/6

(remember in trig, counterclockwise is positive and clockwise is negative)
or
consider the rotations in degrees
31π/6 radians = 930° = 5(360°) + 30° , then
-31π/6 radians = -930° = 5(-360°) - 30°

check with your calculator:
tan 930° = tan30°
tan (-930°) = tan (-30°

oh i see! thank you!

To evaluate the expression arctan(tan(-31π/6), we need to understand the properties of the arctangent and tangent functions.

The tangent function (tan(x)) takes an angle as an input and gives the ratio of the sine and cosine of that angle. It's important to note that the tangent function has a periodicity of π or 180 degrees.

The arctangent function (arctan(x)), on the other hand, takes a ratio as an input and gives the corresponding angle.

In this case, we have arctan(tan(-31π/6)). We'll first simplify the expression inside the tangent function using the periodicity of the tangent function.

Since 31π/6 is greater than 2π, we need to find the angle within the range of 0 to 2π that has the same tangent value. To do this, we can subtract 2π from 31π/6.

31π/6 - 2π = (31π - 12π)/6 = 19π/6

Now, we have arctan(tan(19π/6)). The arctan function will return an angle within a specific range. The range for arctan is (-π/2, π/2).

Since 19π/6 is not within the range of (-π/2, π/2), we need to find an equivalent value within that range. We can subtract π from 19π/6.

19π/6 - π = (19π - 6π)/6 = 13π/6

Now, we have arctan(tan(13π/6)). The angle 13π/6 is within the range of (-π/2, π/2), so we can evaluate this using our knowledge of trigonometric functions.

The tangent of 13π/6 is sqrt(3). Therefore, arctan(sqrt(3)) is the answer.

Hence, the evaluation of the expression arctan(tan(-31π/6)) is arctan(sqrt(3)).