Question:
If
\[ \sin^{-1}x+\sin^{-1}y+\sin^{-1}z+\sin^{-1}t = 2\pi \]
Find \(x^2 + y^2 + z^2 + t^2\).
Concept:
The principal value range of \( \sin^{-1}x \) is:
\[ -\frac{\pi}{2} \leq \sin^{-1}x \leq \frac{\pi}{2} \]
So each term is at most \( \frac{\pi}{2} \).
—Solution:
Maximum possible sum of four terms:
\[ 4 \times \frac{\pi}{2} = 2\pi \]
So equality holds only when:
\[ \sin^{-1}x = \sin^{-1}y = \sin^{-1}z = \sin^{-1}t = \frac{\pi}{2} \]
Thus:
\[ x = y = z = t = 1 \]
Step 2: Compute required value
\[ x^2 + y^2 + z^2 + t^2 = 1 + 1 + 1 + 1 = 4 \]
—Final Answer:
\[ \boxed{4} \]