The expression $\tan\left(\frac{5\pi}{6}\right)$ involves finding the tangent of an angle in radians. Let’s simplify:

Step 1: Locate the angle

$\frac{5\pi}{6}$ is in the second quadrant (since $\pi/2 < 5\pi/6 < \pi$).

Step 2: Use the reference angle

The reference angle is $\pi - \frac{5\pi}{6} = \frac{\pi}{6}$.

Step 3: Determine the sign of $\tan$

In the second quadrant, $\tan$ is negative.

Step 4: Evaluate $\tan\left(\frac{\pi}{6}\right)$

$\tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}.$

Step 5: Apply the sign

Since $\tan\left(\frac{5\pi}{6}\right)$ is negative: $\tan\left(\frac{5\pi}{6}\right) = -\frac{1}{\sqrt{3}}.$

Final Answer:

$\tan\left(\frac{5\pi}{6}\right) = -\frac{1}{\sqrt{3}} \text{ or } -\frac{\sqrt{3}}{3} \text{ (rationalized)}.$

---

Would you like further explanation or related examples?

---

Related Questions:

1. What is $\sin\left(\frac{5\pi}{6}\right)$ and $\cos\left(\frac{5\pi}{6}\right)$?
2. How does the unit circle help in evaluating trigonometric functions?
3. Can we calculate $\tan\left(\frac{5\pi}{6}\right)$ using its Taylor series?
4. What is the general formula for $\tan(n\pi + \theta)$?
5. How does $\tan(\theta)$ behave in different quadrants?

---

Tip:

Remember, tangent is positive in the first and third quadrants, and negative in the second and fourth quadrants. Use the reference angle to simplify calculations!