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To determine which of the given options is equivalent to [tex]\(\tan \left(\frac{5 \pi}{6}\right)\)[/tex], we need to evaluate each trigonometric function step by step.
First, let's recall some properties of the tangent function and the unit circle:
1. Negative Angles and Tangent Function:
[tex]\[ \tan(-\theta) = -\tan(\theta) \][/tex]
2. Tangent Function Periodicity:
[tex]\[ \tan(\theta + \pi) = \tan(\theta) \][/tex]
3. Cotangent Relationship:
[tex]\[ \cot(\theta) = \frac{1}{\tan(\theta)} \][/tex]
Given these properties, let's compare each option to [tex]\(\tan \left(\frac{5 \pi}{6}\right)\)[/tex]:
### Option 1: [tex]\(\tan \left(-\frac{5 \pi}{6}\right)\)[/tex]
By using the property of tangent for negative angles:
[tex]\[ \tan \left(-\frac{5 \pi}{6}\right) = -\tan \left(\frac{5 \pi}{6}\right) \][/tex]
So, [tex]\(\tan \left(-\frac{5 \pi}{6}\right)\)[/tex] is not equivalent to [tex]\(\tan \left(\frac{5 \pi}{6}\right)\)[/tex] since it is the negation.
### Option 2: [tex]\(\tan \left(-\frac{\pi}{6}\right)\)[/tex]
Again, applying the property for negative angles:
[tex]\[ \tan \left(-\frac{\pi}{6}\right) = -\tan \left(\frac{\pi}{6}\right) \][/tex]
We know that [tex]\(\frac{\pi}{6}\)[/tex] lies in the first quadrant, where [tex]\(\tan \left(\frac{\pi}{6}\right) > 0\)[/tex]:
[tex]\[ \tan \left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}} \][/tex]
Thus:
[tex]\[ \tan \left(-\frac{\pi}{6}\right) = -\frac{1}{\sqrt{3}} \][/tex]
To check equivalence, notice that:
[tex]\[ \tan \left(\frac{5 \pi}{6}\right) \text{ is in the second quadrant where } \tan \left(\frac{5 \pi}{6}\right) = -\tan \left(\frac{\pi}{6}\right) = -\frac{1}{\sqrt{3}} \][/tex]
Thus:
[tex]\[ \tan \left(-\frac{\pi}{6}\right) = \tan \left(\frac{5 \pi}{6}\right) \][/tex]
### Option 3: [tex]\(\cot \left(\frac{5 \pi}{6}\right)\)[/tex]
Using the cotangent relationship:
[tex]\[ \cot \left(\frac{5 \pi}{6}\right) = \frac{1}{\tan \left(\frac{5 \pi}{6}\right)} \][/tex]
This is clearly not equivalent to [tex]\(\tan \left(\frac{5 \pi}{6}\right)\)[/tex].
### Option 4: [tex]\(\tan \left(\frac{7 \pi}{6}\right)\)[/tex]
Using the periodicity:
[tex]\[ \tan \left(\frac{7 \pi}{6}\right) = \tan \left(\frac{\pi}{6} + \pi\right) = \tan \left(\frac{\pi}{6}\right) \][/tex]
Where [tex]\(\frac{\pi}{6}\)[/tex] is in the first quadrant, so:
[tex]\[ \tan \left(\frac{7 \pi}{6}\right) = \frac{1}{\sqrt{3}} \][/tex]
This is not equivalent to [tex]\(\tan \left(\frac{5 \pi}{6}\right) = -\frac{1}{\sqrt{3}}\)[/tex].
Therefore, the answer is:
[tex]\[ \tan \left(-\frac{\pi}{6}\right) \][/tex]
So, the option that is equivalent to [tex]\(\tan \left(\frac{5 \pi}{6}\right)\)[/tex] is [tex]\(\tan \left(-\frac{\pi}{6}\right)\)[/tex], which corresponds to option 2.
First, let's recall some properties of the tangent function and the unit circle:
1. Negative Angles and Tangent Function:
[tex]\[ \tan(-\theta) = -\tan(\theta) \][/tex]
2. Tangent Function Periodicity:
[tex]\[ \tan(\theta + \pi) = \tan(\theta) \][/tex]
3. Cotangent Relationship:
[tex]\[ \cot(\theta) = \frac{1}{\tan(\theta)} \][/tex]
Given these properties, let's compare each option to [tex]\(\tan \left(\frac{5 \pi}{6}\right)\)[/tex]:
### Option 1: [tex]\(\tan \left(-\frac{5 \pi}{6}\right)\)[/tex]
By using the property of tangent for negative angles:
[tex]\[ \tan \left(-\frac{5 \pi}{6}\right) = -\tan \left(\frac{5 \pi}{6}\right) \][/tex]
So, [tex]\(\tan \left(-\frac{5 \pi}{6}\right)\)[/tex] is not equivalent to [tex]\(\tan \left(\frac{5 \pi}{6}\right)\)[/tex] since it is the negation.
### Option 2: [tex]\(\tan \left(-\frac{\pi}{6}\right)\)[/tex]
Again, applying the property for negative angles:
[tex]\[ \tan \left(-\frac{\pi}{6}\right) = -\tan \left(\frac{\pi}{6}\right) \][/tex]
We know that [tex]\(\frac{\pi}{6}\)[/tex] lies in the first quadrant, where [tex]\(\tan \left(\frac{\pi}{6}\right) > 0\)[/tex]:
[tex]\[ \tan \left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}} \][/tex]
Thus:
[tex]\[ \tan \left(-\frac{\pi}{6}\right) = -\frac{1}{\sqrt{3}} \][/tex]
To check equivalence, notice that:
[tex]\[ \tan \left(\frac{5 \pi}{6}\right) \text{ is in the second quadrant where } \tan \left(\frac{5 \pi}{6}\right) = -\tan \left(\frac{\pi}{6}\right) = -\frac{1}{\sqrt{3}} \][/tex]
Thus:
[tex]\[ \tan \left(-\frac{\pi}{6}\right) = \tan \left(\frac{5 \pi}{6}\right) \][/tex]
### Option 3: [tex]\(\cot \left(\frac{5 \pi}{6}\right)\)[/tex]
Using the cotangent relationship:
[tex]\[ \cot \left(\frac{5 \pi}{6}\right) = \frac{1}{\tan \left(\frac{5 \pi}{6}\right)} \][/tex]
This is clearly not equivalent to [tex]\(\tan \left(\frac{5 \pi}{6}\right)\)[/tex].
### Option 4: [tex]\(\tan \left(\frac{7 \pi}{6}\right)\)[/tex]
Using the periodicity:
[tex]\[ \tan \left(\frac{7 \pi}{6}\right) = \tan \left(\frac{\pi}{6} + \pi\right) = \tan \left(\frac{\pi}{6}\right) \][/tex]
Where [tex]\(\frac{\pi}{6}\)[/tex] is in the first quadrant, so:
[tex]\[ \tan \left(\frac{7 \pi}{6}\right) = \frac{1}{\sqrt{3}} \][/tex]
This is not equivalent to [tex]\(\tan \left(\frac{5 \pi}{6}\right) = -\frac{1}{\sqrt{3}}\)[/tex].
Therefore, the answer is:
[tex]\[ \tan \left(-\frac{\pi}{6}\right) \][/tex]
So, the option that is equivalent to [tex]\(\tan \left(\frac{5 \pi}{6}\right)\)[/tex] is [tex]\(\tan \left(-\frac{\pi}{6}\right)\)[/tex], which corresponds to option 2.
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