Evaluate the expression. tan − 1 1 \tan^{-1}1 tan − 1 1

this problem, we're asked to evaluate the expression the inverse tangent of 1. Now here we can also think of this as okay the tangent of what angle is going to be equal to 1 and we're trying to find the angle for which that is true. Now I know that my angle has to be in between negative pi over 2 and positive pi over 2 because that is our specified interval when working with the inverse tangent. And I also know that because this is a number, positive one, I am specifically looking in this first quadrant where all of my tangent values are positive. So for which value in this or for which angle in this first quadrant is the tangent positive 1? Well, for pi over 4, I see that my tangent is indeed 1, so that represents my solution, pi over 4 for the inverse tangent of 1. Thanks for watching and I'll see you in the next one.

Evaluate the expression. tan ⁡ − 1 1 \tan^{-1}1 tan − 1 1

− π 4 -\frac{\pi}{4} − 4 π

5. Inverse Trigonometric Functions and Basic Trigonometric Equations

Inverse Sine, Cosine, & Tangent

Trigonometry

5. Inverse Trigonometric Functions and Basic Trigonometric Equations

Inverse Sine, Cosine, & Tangent

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Multiple Choice

Evaluate the expression.
$\tan^{-1}1$

$0$

$\frac{\pi}{4}$

$\frac{\pi}{2}$

$-\frac{\pi}{4}$

Verified step by step guidance

Understand that $ \tan^{-1}(x) $ is the inverse tangent function, which gives the angle whose tangent is $ x $.

Recognize that $ \tan^{-1}(1) $ asks for the angle whose tangent is 1.

Recall that the tangent of $ \frac{\pi}{4} $ is 1, because $ \tan(\frac{\pi}{4}) = \frac{\sin(\frac{\pi}{4})}{\cos(\frac{\pi}{4})} = \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = 1 $.

Therefore, $ \tan^{-1}(1) = \frac{\pi}{4} $, since $ \frac{\pi}{4} $ is the principal value of the inverse tangent function for this input.

Conclude that the correct answer is $ \frac{\pi}{4} $, as it is the angle in the range $-\frac{\pi}{2}, \frac{\pi}{2}$ that satisfies the condition.