The problem visible in the image involves finding $\tan \left( \frac{11\pi}{6} \right)$.

Let's break it down step by step:

Step 1: Convert $\frac{11\pi}{6}$ to degrees

We know that $\pi$ radians equals $180^\circ$. Thus, $\frac{11\pi}{6} \times \frac{180^\circ}{\pi} = \frac{11 \times 180^\circ}{6} = 330^\circ.$ So, $\frac{11\pi}{6} = 330^\circ$.

Step 2: Find the reference angle

The reference angle for $330^\circ$ can be found by subtracting $330^\circ$ from $360^\circ$: $360^\circ - 330^\circ = 30^\circ.$ Thus, the reference angle is $30^\circ$.

Step 3: Determine the quadrant and sign of the tangent

Since $330^\circ$ lies in the fourth quadrant, and the tangent function is negative in this quadrant, $\tan(330^\circ)$ will be negative.

Step 4: Evaluate $\tan(30^\circ)$

We know that: $\tan(30^\circ) = \frac{1}{\sqrt{3}}.$

Step 5: Apply the sign from the quadrant

Therefore, $\tan(330^\circ) = -\frac{1}{\sqrt{3}}$.

To rationalize the denominator: $\tan(330^\circ) = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3}.$

Final Answer:

$\tan \left( \frac{11\pi}{6} \right) = -\frac{\sqrt{3}}{3}.$

It appears the answer written in the image is correct, but the sign should be negative. Would you like any further details or clarifications on this?

Here are five related questions you can explore:

1. What is the tangent of $\frac{7\pi}{6}$?
2. How do you convert radians to degrees and vice versa?
3. What are the properties of trigonometric functions in different quadrants?
4. How do you rationalize a denominator?
5. What is the tangent of $\frac{5\pi}{3}$?

Tip: Always determine the correct quadrant when solving trigonometric functions, as the sign of the result depends on the quadrant.