To find $\cot(-\frac{5\pi}{6})$, let's go step-by-step:

Step 1: Understanding the angle

The given angle is $-\frac{5\pi}{6}$. In terms of radians, we need to simplify this by understanding its position on the unit circle.

1. Since $\frac{\pi}{6}$ radians is 30°, we have $\frac{5\pi}{6} = 150^\circ$.
2. Therefore, $-\frac{5\pi}{6}$ corresponds to an angle measured clockwise by 150°.

Step 2: Finding the reference angle

The reference angle is the positive acute angle formed with the x-axis.

• The reference angle for $-\frac{5\pi}{6}$ is $\pi - \frac{5\pi}{6} = \frac{\pi}{6}$.
• The reference angle is $\frac{\pi}{6}$, or 30°.

Step 3: Determine the quadrant

$-\frac{5\pi}{6}$ falls in the third quadrant since:

• Starting from 0 radians (on the positive x-axis), rotating clockwise by $150^\circ$ moves us into the third quadrant.

In the third quadrant, both sine and cosine are negative, and therefore the cotangent is positive.

Step 4: Cotangent of reference angle

We use the reference angle $\frac{\pi}{6}$ (which corresponds to 30°). The value of $\cot(\frac{\pi}{6})$ is:

$\cot(\frac{\pi}{6}) = \frac{1}{\tan(\frac{\pi}{6})} = \frac{1}{\frac{1}{\sqrt{3}}} = \sqrt{3}$

Since the cotangent is positive in the third quadrant, we have:

$\cot\left(-\frac{5\pi}{6}\right) = \sqrt{3}$

Final Answer:

$\boxed{\cot\left(-\frac{5\pi}{6}\right) = \sqrt{3}}$

Would you like further explanation or details on any of the steps?

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Here are 5 related questions for further exploration:

1. What is $\cot\left(\frac{5\pi}{6}\right)$?
2. How do you find the cotangent of any angle in the second quadrant?
3. What is $\sin\left(-\frac{5\pi}{6}\right)$?
4. How do you find the reference angle of a negative angle?
5. What is the exact value of $\tan\left(-\frac{7\pi}{4}\right)$?

Tip: Always verify the quadrant of your angle to determine the sign of trigonometric functions!