Find x+y+z when sin⁻¹x + sin⁻¹y + sin⁻¹z = 3π/2

Find the Value of x + y + z

Given:
\[ \sin^{-1}x + \sin^{-1}y + \sin^{-1}z = \frac{3\pi}{2} \]
Concept Used:
The principal value range of inverse sine is: \[ \sin^{-1}t \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \]

Maximum value of each term = \( \frac{\pi}{2} \)
Step 1:
Since the sum of three inverse sine functions is: \[ \frac{3\pi}{2} \] this is the maximum possible sum.
Step 2:
This happens only when: \[ \sin^{-1}x = \sin^{-1}y = \sin^{-1}z = \frac{\pi}{2} \]
Step 3:
\[ \sin^{-1}x = \frac{\pi}{2} \Rightarrow x = 1 \] \[ \sin^{-1}y = \frac{\pi}{2} \Rightarrow y = 1 \] \[ \sin^{-1}z = \frac{\pi}{2} \Rightarrow z = 1 \]
Step 4: Required Value
\[ x + y + z = 1 + 1 + 1 = 3 \]
Final Answer: \[ 3 \]
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