To rewrite $\tan\left(\sin^{-1}\left(x\right)\right)$ as an algebraic expression in $x$, follow these steps:

Step 1: Understand the components

The expression involves:

• $\sin^{-1}(x)$, which is the inverse sine function. Let $\theta = \sin^{-1}(x)$, so $\sin(\theta) = x$.
• The goal is to find $\tan(\theta)$ in terms of $x$.

Step 2: Use the Pythagorean identity

From the definition of the sine function: $\sin(\theta) = x \implies \frac{\text{opposite}}{\text{hypotenuse}} = x.$ Let the hypotenuse be 1 and the opposite side be $x$. Using the Pythagorean theorem, the adjacent side is: $\sqrt{1 - x^2}.$

Step 3: Express $\tan(\theta)$

The tangent function is defined as: $\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}.$ Substitute the values: $\tan(\theta) = \frac{x}{\sqrt{1 - x^2}}.$

Final Answer:

$\tan\left(\sin^{-1}(x)\right) = \frac{x}{\sqrt{1 - x^2}}.$

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Related Questions:

1. How do we find similar algebraic expressions for $\sec\left(\sin^{-1}(x)\right)$?
2. Can we rewrite $\cos\left(\sin^{-1}(x)\right)$ as an algebraic expression?
3. What is the domain of $\tan\left(\sin^{-1}(x)\right)$?
4. How does the unit circle relate to this derivation?
5. How can trigonometric identities simplify inverse trigonometric functions further?

Tip:

Always check the domain restrictions for the square root to ensure the expression is valid.