We are tasked to find the tangent of an angle $\theta$ given that $\cos(\theta) = -\frac{1}{\sqrt{6}}$ and the angle lies in Quadrant III.

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Step 1: Use the Pythagorean Identity to Find $\sin(\theta)$

The Pythagorean identity is:

$\sin^2(\theta) + \cos^2(\theta) = 1.$

Substitute $\cos(\theta) = -\frac{1}{\sqrt{6}}$:

$\sin^2(\theta) + \left(-\frac{1}{\sqrt{6}}\right)^2 = 1.$

Simplify:

$\sin^2(\theta) + \frac{1}{6} = 1.$

$\sin^2(\theta) = 1 - \frac{1}{6}.$

$\sin^2(\theta) = \frac{6}{6} - \frac{1}{6} = \frac{5}{6}.$

Thus:

$\sin(\theta) = \pm \sqrt{\frac{5}{6}}.$

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Step 2: Determine the Sign of $\sin(\theta)$

In Quadrant III, both sine and cosine are negative. Therefore:

$\sin(\theta) = -\sqrt{\frac{5}{6}}.$

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Step 3: Use the Definition of Tangent

The tangent of an angle is defined as:

$\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}.$

Substitute $\sin(\theta) = -\sqrt{\frac{5}{6}}$ and $\cos(\theta) = -\frac{1}{\sqrt{6}}$:

$\tan(\theta) = \frac{-\sqrt{\frac{5}{6}}}{-\frac{1}{\sqrt{6}}}.$

Simplify the signs:

$\tan(\theta) = \frac{\sqrt{\frac{5}{6}}}{\frac{1}{\sqrt{6}}}.$

Simplify the fraction:

$\tan(\theta) = \sqrt{\frac{5}{6}} \cdot \sqrt{6}.$

$\tan(\theta) = \sqrt{5}.$

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Final Answer:

$\tan(\theta) = \sqrt{5}.$

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Do you have any questions, or would you like a detailed explanation of any step? Here are five related questions for further exploration:

1. How do the signs of trigonometric functions vary across the four quadrants?
2. Can you use the reciprocal identities (secant, cosecant) to verify the values found here?
3. What would the tangent be if $\theta$ were in Quadrant II instead?
4. How is the unit circle related to the Pythagorean identity?
5. How does knowing the value of $\tan(\theta)$ help in solving triangles?

Tip: Always determine the signs of sine and cosine using the quadrant information to ensure the correct value of the tangent!