To solve the given problem:

We are tasked to rewrite

$\cos\left(\cos^{-1}(x) + \sin^{-1}(x)\right)$

as an algebraic expression in terms of $x$.

Step-by-Step Solution:

Step 1: Recall the properties of inverse trigonometric functions

1. From the property of inverse cosine and sine: $\cos^{-1}(x) + \sin^{-1}(x) = \frac{\pi}{2}, \text{ for all } -1 \leq x \leq 1.$

Step 2: Simplify the expression

Using the property above, substitute: $\cos^{-1}(x) + \sin^{-1}(x) = \frac{\pi}{2}.$

Thus, the original expression becomes: $\cos\left(\cos^{-1}(x) + \sin^{-1}(x)\right) = \cos\left(\frac{\pi}{2}\right).$

Step 3: Evaluate $\cos\left(\frac{\pi}{2}\right)$

The cosine of $\frac{\pi}{2}$ is: $\cos\left(\frac{\pi}{2}\right) = 0.$

Final Answer:

$\cos\left(\cos^{-1}(x) + \sin^{-1}(x)\right) = 0.$

Would you like a detailed explanation or further exploration of this concept?

Related Questions:

1. What is the domain of $\cos^{-1}(x)$ and $\sin^{-1}(x)$?
2. How does $\cos\left(\cos^{-1}(x)\right)$ simplify for different values of $x$?
3. Can $\sin^{-1}(x) + \cos^{-1}(x)$ be greater than $\pi$?
4. What is the general range of $\cos\left(\cos^{-1}(x) + \sin^{-1}(x)\right)$?
5. How would this result change if a different trigonometric function (e.g., sine) were used?

Tip:

Always verify inverse trigonometric identities when combining them, as they often simplify complex expressions!