1. **Stating the problem:** Find the value of $\sin(\tan^{-1}(x))$ or understand the relationship between sine and tangent functions. 2. **Formula and rules:** Recall that $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$ and $\tan^{-1}(x)$ is the angle whose tangent is $x$. 3. **Intermediate work:** Let $\theta = \tan^{-1}(x)$, so $\tan(\theta) = x = \frac{\sin(\theta)}{\cos(\theta)}$. 4. From the right triangle definition, if $\tan(\theta) = x = \frac{\text{opposite}}{\text{adjacent}}$, set opposite side = $x$ and adjacent side = 1. 5. Then hypotenuse $= \sqrt{x^2 + 1}$. 6. Therefore, $\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{x}{\sqrt{x^2 + 1}}$. 7. **Final answer:** $$\sin(\tan^{-1}(x)) = \frac{x}{\sqrt{x^2 + 1}}$$ This shows how to express sine of an inverse tangent in terms of $x$.