If `sin^(-1)x+sin^(-1)y+sin^(-1)z=(3pi)/2` , then write the value of `x+y+z` .

Explanation

To solve the equation \( \sin^{-1}x + \sin^{-1}y + \sin^{-1}z = \frac{3\pi}{2} \) and find the value of \( x + y + z \), we can follow these steps:

### Step 1: Understand the range of the inverse sine function
The function \( \sin^{-1}x \) (or arcsin) has a range of \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \). Therefore, the maximum value of \( \sin^{-1}x \) is \( \frac{\pi}{2} \), which occurs when \( x = 1 \).

### Step 2: Analyze the equation
Given that \( \sin^{-1}x + \sin^{-1}y + \sin^{-1}z = \frac{3\pi}{2} \), we can deduce that each term \( \sin^{-1}x \), \( \sin^{-1}y \), and \( \sin^{-1}z \) must be at their maximum value of \( \frac{\pi}{2} \) to sum to \( \frac{3\pi}{2} \).

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