To evaluate $\tan(\arcsin(x))$, we start by setting $\theta = \arcsin(x)$. This tells us that:

$\sin(\theta) = x$

Since the range of $\arcsin(x)$ is $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$, $\theta$ is in the first or fourth quadrant. By definition of the sine function, we have:

$\frac{\text{opposite}}{\text{hypotenuse}} = \frac{x}{1}$

To find $\tan(\theta)$, we need the ratio $\frac{\text{opposite}}{\text{adjacent}}$. Using the Pythagorean identity for a right triangle:

$\text{adjacent} = \sqrt{\text{hypotenuse}^2 - \text{opposite}^2} = \sqrt{1 - x^2}$

Thus, $\tan(\theta)$ becomes:

$\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{x}{\sqrt{1 - x^2}}$

Therefore:

$\tan(\arcsin(x)) = \frac{x}{\sqrt{1 - x^2}}$

Would you like me to expand on this or apply it to a specific value?

Related Questions:

1. What is $\tan(\arccos(x))$?
2. How do you derive $\tan(\arctan(x))$?
3. What is the range and domain of $\arcsin(x)$?
4. What is the derivative of $\tan(\arcsin(x))$?
5. How do trigonometric identities help simplify expressions like these?

Tip: For simplifying trigonometric expressions with inverse functions, drawing a right triangle often helps in visualizing and finding other trigonometric ratios.