To evaluate $\sin(\tan^{-1}(\sqrt{3}))$, let's proceed step-by-step.

---

Step 1: Let $\theta = \tan^{-1}(\sqrt{3})$

This implies:

$\tan(\theta) = \sqrt{3}.$

From the definition of tangent:

$\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}},$

we can assign:

$\text{opposite} = \sqrt{3}, \quad \text{adjacent} = 1.$

---

Step 2: Find the hypotenuse

Using the Pythagorean theorem:

$\text{hypotenuse} = \sqrt{(\text{opposite})^2 + (\text{adjacent})^2} = \sqrt{(\sqrt{3})^2 + (1)^2} = \sqrt{3 + 1} = 2.$

Thus, the sides of the right triangle are:

$\text{opposite} = \sqrt{3}, \quad \text{adjacent} = 1, \quad \text{hypotenuse} = 2.$

---

Step 3: Find $\sin(\theta)$

From the definition of sine:

$\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}.$

Substituting the values:

$\sin(\theta) = \frac{\sqrt{3}}{2}.$

---

Final Answer:

$\sin(\tan^{-1}(\sqrt{3})) = \frac{\sqrt{3}}{2}.$

Let me know if you'd like a more detailed explanation!

---

Expanding Questions:

1. How is the Pythagorean theorem used to find the hypotenuse in trigonometric problems?
2. What is the relationship between inverse trigonometric functions and right triangles?
3. Can this method be generalized for any $\tan^{-1}(x)$?
4. How does this approach change if $x$ is negative (e.g., $\tan^{-1}(-\sqrt{3})$)?
5. What are the possible domains and ranges of $\sin(\tan^{-1}(x))$?

---

Tip: Understanding the relationship between trigonometric functions and right triangles helps to solve complex expressions efficiently.