Evaluate the expression. tan − 1 0 \tan^{-1}0 tan − 1 0

this problem, we're asked to evaluate the expression the inverse tangent of 0. So here we're looking for an angle for which the tangent is equal to 0. Now for the inverse tangent, I know that my values can only be in between negative pi over 2 and positive pi over 2. And because this is neither a positive or negative value that we're taking the inverse tangent of, I also might recognize that it's right in between my 2 quadrants. So here looking at my value of 0, this is one of my quadrantal angles for which I know the tangent is equal to 0. So here the angle that I'm looking for is simply 0 and that represents my solution. Thanks for watching, and I'll see you in the next one.

Evaluate the expression. tan ⁡ − 1 0 \tan^{-1}0 tan − 1 0

− π 2 -\frac{\pi}{2} − 2 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.
$\tan^{-1}0$

$0$

$\frac{\pi}{2}$

$\pi$

$-\frac{\pi}{2}$

Verified step by step guidance

Understand the problem: We need to evaluate the expression $ \tan^{-1}(0) $. This involves finding the angle whose tangent is 0.

Recall the definition of the inverse tangent function: $ \tan^{-1}(x) $ gives the angle $ \theta $ such that $ \tan(\theta) = x $.

Identify the angle: The tangent of an angle is 0 at $ \theta = 0 $ and $ \theta = \pi $ (or any multiple of $ \pi $), but $ \tan^{-1}(0) $ specifically refers to the principal value, which is $ \theta = 0 $.

Consider the range of $ \tan^{-1}(x) $: The principal value of $ \tan^{-1}(x) $ is typically in the range $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2} $. Within this range, $ \tan^{-1}(0) = 0 $.

Conclude the evaluation: Since $ \tan^{-1}(0) = 0 $, the expression evaluates to $ 0 $.