1. **State the problem:** Calculate $\tan\left(\frac{92\pi}{3}\right)$.\n\n2. **Recall the periodicity of tangent:** The tangent function has a period of $\pi$, meaning $\tan(x) = \tan(x + k\pi)$ for any integer $k$.\n\n3. **Reduce the angle modulo $\pi$:** We want to find an equivalent angle $\theta$ such that $\theta = \frac{92\pi}{3} - k\pi$ and $0 \leq \theta < \pi$.\n\n4. Calculate $k$:\n$$k = \left\lfloor \frac{92}{3} \right\rfloor = 30$$\n\n5. Subtract $30\pi$ from $\frac{92\pi}{3}$:\n$$\frac{92\pi}{3} - 30\pi = \frac{92\pi}{3} - \frac{90\pi}{3} = \frac{2\pi}{3}$$\n\n6. So, $\tan\left(\frac{92\pi}{3}\right) = \tan\left(\frac{2\pi}{3}\right)$.\n\n7. **Evaluate $\tan\left(\frac{2\pi}{3}\right)$:**\nRecall that $\tan\left(\pi - x\right) = -\tan(x)$, so\n$$\tan\left(\frac{2\pi}{3}\right) = \tan\left(\pi - \frac{\pi}{3}\right) = -\tan\left(\frac{\pi}{3}\right) = -\sqrt{3}$$\n\n**Final answer:**\n$$\tan\left(\frac{92\pi}{3}\right) = -\sqrt{3}$$