Evaluate the expression.tan−1(−33)\tan^{-1}\left(-\frac{\sqrt3}{3}\right)tan−1(−33)

− π 6 -\frac{\pi}{6} − 6 π

Evaluate the expression. tan − 1 ( − 3 3 ) \tan^{-1}\left(-\frac{\sqrt3}{3}\right) tan − 1 ( − 3 3 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z"> )

this problem, we're asked to evaluate the expression of the inverse tangent of negative root 3 over 3. So we're looking for the angle for which the tangent is equal to negative root 3 over 3. Now, because this is a negative value, I know that I'm specifically looking for an angle within my interval that is in quadrant 4 because that's where all of my tangent values are negative within my interval from negative pi over 2 to positive pi over 2. So in this 4th quadrant, which one of these angles is going to give me a tangent of negative root 3 over 3? Well, I know that negative pi over 6 is a reference angle with pi over 6. So I know that the tangent here is negative root 3 over 3 and that represents my solution, negative pi over 6. Remember, this is specifically not 11 pi over 6 because that would not be within our specified interval from negative pi over 2 to positive pi over 2. So my solution here is negative pi over 6 and we're done here. Thanks for watching and I'll see you in the next one.

Evaluate the expression. tan ⁡ − 1 ( − 3 3 ) \tan^{-1}\left(-\frac{\sqrt3}{3}\right) tan − 1 ( − 3 3 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> )

− π 6 -\frac{\pi}{6} − 6 π

− π 3 -\frac{\pi}{3} − 3 π

11. Inverse Trigonometric Functions and Basic Trig Equations

Inverse Sine, Cosine, & Tangent

Precalculus

11. Inverse Trigonometric Functions and Basic Trig Equations

Inverse Sine, Cosine, & Tangent

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Multiple Choice

Evaluate the expression.
$\tan^{-1}\left(-\frac{\sqrt3}{3}\right)$

$\frac{\pi}{6}$

$\frac{\pi}{3}$

$-\frac{\pi}{6}$

$-\frac{\pi}{3}$

Verified step by step guidance

Understand that the problem involves evaluating the inverse tangent function, $ \tan^{-1} $, which gives the angle whose tangent is the given value.

Recognize that $ \tan^{-1}(-\frac{\sqrt{3}}{3}) $ asks for the angle whose tangent is $ -\frac{\sqrt{3}}{3} $.

Recall that the tangent of $ \frac{\pi}{6} $ is $ \frac{1}{\sqrt{3}} $, which is equivalent to $ \frac{\sqrt{3}}{3} $. Therefore, $ \tan^{-1}(\frac{\sqrt{3}}{3}) = \frac{\pi}{6} $.

Since the tangent function is negative in the fourth quadrant, the angle corresponding to $ \tan^{-1}(-\frac{\sqrt{3}}{3}) $ is $ -\frac{\pi}{6} $.

Conclude that the correct angle is $ -\frac{\pi}{6} $, as it is the angle in the fourth quadrant with the same reference angle as $ \frac{\pi}{6} $.

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