To simplify the expression $\sqrt{1+x} \cdot \sqrt{1-x}$, we use the property of square roots:

$\sqrt{a} \cdot \sqrt{b} = \sqrt{a \cdot b}$

Step-by-Step Simplification:

1. Apply the square root product property: $\sqrt{1+x} \cdot \sqrt{1-x} = \sqrt{(1+x)(1-x)}$

2. Simplify the product inside the square root: $(1+x)(1-x) = 1^2 - x^2 = 1 - x^2 \quad \text{(difference of squares)}.$

3. Substitute back: $\sqrt{1+x} \cdot \sqrt{1-x} = \sqrt{1-x^2}.$

Final Answer:

$\sqrt{1+x} \cdot \sqrt{1-x} = \sqrt{1-x^2}.$

Would you like further clarification or examples related to this?

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Here are 5 related questions to explore:

1. What is the domain of the function $\sqrt{1 - x^2}$?
2. How would you simplify $\sqrt{a^2 - b^2}$?
3. How does the difference of squares apply to more complex algebraic expressions?
4. What are the implications if $x > 1$ or $x < -1$ in the expression $\sqrt{1-x^2}$?
5. How do you handle square roots of negative expressions in this context?

Tip: Always verify the domain of the variables to ensure the square root is defined!