The problem is to find all numbers $t$ such that:

$\cos^{-1}(t) = \sin^{-1}(t).$

Solution:

1. Understanding the relationship: The domains of $\cos^{-1}(t)$ and $\sin^{-1}(t)$ are: $-1 \leq t \leq 1.$

   The values of the inverse cosine and inverse sine functions satisfy: $\cos^{-1}(t) + \sin^{-1}(t) = \frac{\pi}{2}.$

2. Set up the equation: If $\cos^{-1}(t) = \sin^{-1}(t)$, substitute this into the above identity: $\cos^{-1}(t) + \cos^{-1}(t) = \frac{\pi}{2}.$

3. Solve for $\cos^{-1}(t)$: $2 \cos^{-1}(t) = \frac{\pi}{2}.$ $\cos^{-1}(t) = \frac{\pi}{4}.$

4. Find $t$: Using the definition of the inverse cosine function: $\cos^{-1}(t) = \frac{\pi}{4} \implies t = \cos\left(\frac{\pi}{4}\right).$

   From trigonometric values: $t = \frac{\sqrt{2}}{2}.$

Final Answer:

The only solution is: $t = \frac{\sqrt{2}}{2}.$

Let me know if you want further clarification or details!

Related Questions:

1. What is the domain and range of $\cos^{-1}(x)$ and $\sin^{-1}(x)$?
2. How do you derive the identity $\cos^{-1}(x) + \sin^{-1}(x) = \frac{\pi}{2}$?
3. What is the geometric meaning of $\cos^{-1}(x) = \sin^{-1}(x)$?
4. Can this equation have any solutions outside $-1 \leq t \leq 1$? Why or why not?
5. What happens if $\cos^{-1}(x) = 2\sin^{-1}(x)$? How would the solution change?

Tip:

When working with inverse trigonometric equations, always verify that solutions lie within the domains of the functions involved.