To find \( \tan(\sin^{-1}(x)) \), we can proceed with the following steps:

### Step-by-step solution:

1. **Let \( \theta = \sin^{-1}(x) \)**, so that \( \sin(\theta) = x \).
2. Now, we need to find \( \tan(\theta) \). Using the identity for the tangent in terms of sine and cosine:
   \[
   \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}
   \]
3. Since \( \sin(\theta) = x \), we now need to find \( \cos(\theta) \). Use the Pythagorean identity for sine and cosine:
   \[
   \sin^2(\theta) + \cos^2(\theta) = 1
   \]
   Substituting \( \sin(\theta) = x \):
   \[
   x^2 + \cos^2(\theta) = 1
   \]
   Solving for \( \cos(\theta) \):
   \[
   \cos(\theta) = \sqrt{1 - x^2}
   \]
4. Now, substitute the values of \( \sin(\theta) \) and \( \cos(\theta) \) into the tangent expression:
   \[
   \tan(\theta) = \frac{x}{\sqrt{1 - x^2}}
   \]

Thus, \( \tan(\sin^{-1}(x)) = \frac{x}{\sqrt{1 - x^2}} \).

### Final answer:
\[
\tan(\sin^{-1}(x)) = \frac{x}{\sqrt{1 - x^2}}
\]

Do you need further details or have any questions?

Here are 5 related questions for further exploration:
1. What is \( \tan(\cos^{-1}(x)) \)?
2. How would you find \( \tan(\sin^{-1}(x) + \cos^{-1}(x)) \)?
3. What is the range of values for which \( \tan(\sin^{-1}(x)) \) is valid?
4. How does the identity change for inverse trigonometric functions of complex numbers?
5. Can you derive \( \tan(\sin^{-1}(2x)) \)?

**Tip**: Always check the domain of the function when dealing with inverse trigonometric functions.

Explore how to find tan(sin^(-1)(x)) with a detailed step-by-step solution. This problem covers trigonometric identities, including the Pythagorean identity and the definition of tangent in terms of sine and cosine. Ideal for students in Grades 10-12.

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