Determine the domain of the argument of the tangent function.

The function is $f(x) = \tan\left(\frac{\pi}{\sqrt{x+1}+4}\right)$.
For the expression $\sqrt{x+1}$ to be defined, we must have $x+1 \ge 0$, which implies $x \ge -1$.
The denominator $\sqrt{x+1}+4$ is always positive since $\sqrt{x+1} \ge 0$, so it will never be zero.
Thus, the domain for $x$ is $[-1, \infty)$.

Let $g(x) = \frac{\pi}{\sqrt{x+1}+4}$.
We need to find the range of $g(x)$ for $x \in [-1, \infty)$.

• When $x = -1$: $$g(-1) = \frac{\pi}{\sqrt{-1+1}+4} = \frac{\pi}{\sqrt{0}+4} = \frac{\pi}{0+4} = \frac{\pi}{4}$$
• As $x \to \infty$: As $x$ increases, $\sqrt{x+1}$ increases, so $\sqrt{x+1}+4$ increases and approaches $\infty$. Therefore, $g(x) = \frac{\pi}{\sqrt{x+1}+4}$ approaches 0. Since $\sqrt{x+1}+4$ is an increasing function for $x \ge -1$, the fraction $\frac{\pi}{\sqrt{x+1}+4}$ is a decreasing function. So, the range of the argument $\theta = \frac{\pi}{\sqrt{x+1}+4}$ is $(0, \frac{\pi}{4}]$.

We need to find the range of $\tan(\theta)$ for $\theta \in (0, \frac{\pi}{4}]$.
The tangent function is strictly increasing on the interval $(0, \frac{\pi}{4}]$.

• As $\theta \to 0^+$: $$\lim_{\theta \to 0^+} \tan(\theta) = \tan(0) = 0$$
• When $\theta = \frac{\pi}{4}$: $$\tan\left(\frac{\pi}{4}\right) = 1$$

Therefore, the range of $f(x)$ is $(0, 1]$.

Comparing this with the given options:

1. $[-1,1]$
2. $(0,1]$
3. $[-1,\infty)$

The correct option is 2.