To simplify $\tan(\sin^{-1}(x))$, follow these steps:

1. Let $\theta = \sin^{-1}(x)$, which implies that $\sin(\theta) = x$.

   • Since $\theta = \sin^{-1}(x)$, it means $\theta$ is an angle whose sine is $x$, or $\sin(\theta) = x$.
   • This gives us the identity $\sin(\theta) = \frac{x}{1}$, meaning the opposite side is $x$ and the hypotenuse is $1$ in a right triangle.

2. Use the Pythagorean theorem to find the adjacent side.

   • The Pythagorean theorem states $\text{hypotenuse}^2 = \text{opposite}^2 + \text{adjacent}^2$.
   • Therefore, $1^2 = x^2 + \text{adjacent}^2$, so $\text{adjacent} = \sqrt{1 - x^2}$.

3. Now, the tangent of $\theta$ is: $\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{x}{\sqrt{1 - x^2}}.$

Thus, the simplified form of $\tan(\sin^{-1}(x))$ is: $\tan(\sin^{-1}(x)) = \frac{x}{\sqrt{1 - x^2}}.$

Would you like further details or have any questions?

Here are 5 related questions to expand this:

1. How do you simplify $\tan(\cos^{-1}(x))$?
2. What is the domain of $\tan(\sin^{-1}(x))$?
3. Can you find $\sin(\tan^{-1}(x))$?
4. What are the geometric interpretations of inverse trigonometric functions?
5. How do you simplify $\tan(\sec^{-1}(x))$?

Tip: Understanding the triangle relationships when dealing with inverse trigonometric functions helps in simplifying expressions quickly.