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To rewrite \( \tan(\sin^{-1}(x) + \cos^{-1}(y)) \) in terms of \( x \) and \( y \), we can proceed as follows:

### Step-by-Step Solution

1. **Using the Identity for the Sum of Inverse Functions:**

   We know that for any angle \( \theta \),
   \[
   \sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2}.
   \]
   However, in this case, we are dealing with two different variables \( x \) and \( y \). So we cannot directly use this identity, but it gives us a way to relate the angles of \( \sin^{-1}(x) \) and \( \cos^{-1}(y) \).

2. **Set Angles for Clarity:**

   Let:
   \[
   \alpha = \sin^{-1}(x) \quad \text{and} \quad \beta = \cos^{-1}(y).
   \]
   Then we want to find \( \tan(\alpha + \beta) \).

3. **Using the Tangent Sum Formula:**

   Recall that:
   \[
   \tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}.
   \]

4. **Calculate \( \tan(\alpha) \):**

   Since \( \alpha = \sin^{-1}(x) \), we know:
   \[
   \sin \alpha = x.
   \]
   Using the Pythagorean identity \( \sin^2 \alpha + \cos^2 \alpha = 1 \), we find:
   \[
   \cos \alpha = \sqrt{1 - x^2}.
   \]
   Therefore:
   \[
   \tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{x}{\sqrt{1 - x^2}}.
   \]

5. **Calculate \( \tan(\beta) \):**

   Since \( \beta = \cos^{-1}(y) \), we know:
   \[
   \cos \beta = y.
   \]
   Again using the Pythagorean identity, we find:
   \[
   \sin \beta = \sqrt{1 - y^2}.
   \]
   Therefore:
   \[
   \tan \beta = \frac{\sin \beta}{\cos \beta} = \frac{\sqrt{1 - y^2}}{y}.
   \]

6. **Substitute into the Tangent Sum Formula:**

   Now we substitute \( \tan \alpha = \frac{x}{\sqrt{1 - x^2}} \) and \( \tan \beta = \frac{\sqrt{1 - y^2}}{y} \) into the formula:
   \[
   \tan(\alpha + \beta) = \frac{\frac{x}{\sqrt{1 - x^2}} + \frac{\sqrt{1 - y^2}}{y}}{1 - \frac{x}{\sqrt{1 - x^2}} \cdot \frac{\sqrt{1 - y^2}}{y}}.
   \]

7. **Simplify the Expression:**

   Expanding the numerator and the denominator separately:
   - **Numerator:**
     \[
     \frac{x}{\sqrt{1 - x^2}} + \frac{\sqrt{1 - y^2}}{y}.
     \]
   - **Denominator:**
     \[
     1 - \frac{x \cdot \sqrt{1 - y^2}}{y \cdot \sqrt{1 - x^2}}.
     \]

   Combining these terms might not lead to a simpler form without specific values for \( x \) and \( y \), so this is the most simplified answer in terms of \( x \) and \( y \).

### Final Answer:
\[
\tan(\sin^{-1}(x) + \cos^{-1}(y)) = \frac{\frac{x}{\sqrt{1 - x^2}} + \frac{\sqrt{1 - y^2}}{y}}{1 - \frac{x \cdot \sqrt{1 - y^2}}{y \cdot \sqrt{1 - x^2}}}.
\]

Would you like a more detailed breakdown of any steps, or have any questions?

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### Related Questions
1. How would you simplify \( \tan(\sin^{-1}(x) + \cos^{-1}(y)) \) if given specific values for \( x \) and \( y \)?
2. What are the ranges of \( x \) and \( y \) for \( \sin^{-1}(x) \) and \( \cos^{-1}(y) \) to be defined?
3. Can we apply this method to other trigonometric expressions with different inverse trigonometric functions?
4. How does the tangent sum formula work in cases of complex angles or values?
5. What are some applications of the tangent sum formula in real-world problems?

### Tip
Always remember that the tangent sum formula, \( \tan(\alpha + \beta) \), is a powerful tool for combining angles, especially when dealing with trigonometric expressions in different terms.

Write the given expression in terms of x and y only.
\( \tan(\sin^{-1}(x) + \cos^{-1}(y)) \)

Explore how to simplify \( \tan(\sin^{-1}(x) + \cos^{-1}(y)) \) using trigonometric identities. This problem involves trigonometry concepts such as the tangent sum formula and Pythagorean identity, suitable for advanced high school math students.

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