The value of $\tan\left(\frac{\pi}{6}\right)$ is:

$\tan\left(\frac{\pi}{6}\right) = \frac{\sin\left(\frac{\pi}{6}\right)}{\cos\left(\frac{\pi}{6}\right)}$

From trigonometric values:

• $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$
• $\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$

Substitute these values into the tangent formula: $\tan\left(\frac{\pi}{6}\right) = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3} \quad \text{(after rationalizing the denominator)}.$

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

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Related Questions:

1. What is the exact value of $\tan(\pi/4)$?
2. How can you derive $\tan(x)$ from the unit circle?
3. What are the properties of the tangent function?
4. What is the graph of $\tan(x)$ and its periodicity?
5. How is the tangent function used in solving triangles?

Tip:

Memorizing key trigonometric values like $\sin, \cos,$ and $\tan$ for $\pi/6, \pi/4,$ and $\pi/3$ can simplify solving many trigonometric problems.