If `sin^(-1) x+sin^(-1)y+sin^(-1)z=(3pi)/(2)`, then

Options

A. `x^(100)+y^(100)+z^(100)- (9)/(x^(101)+y^(101)+z^(101))=0`
B. `x^(22)+y^(42)+z^(62)-z^(220)-y^(420)-z^(620)=0`
C. `x^(50)+y^(25)+z^(5)=0`
D. `x^(2008)+y^(2008)+z^(2008)/((xyz)^(2009))=0`

Answer

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

### Step 1: Analyze the Range of \( \sin^{-1} \)
The function \( \sin^{-1} x \) (inverse sine) has a range of \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \). Therefore, the maximum value of \( \sin^{-1} x + \sin^{-1} y + \sin^{-1} z \) occurs when each of \( \sin^{-1} x \), \( \sin^{-1} y \), and \( \sin^{-1} z \) is at its maximum, which is \( \frac{\pi}{2} \).

### Step 2: Calculate the Maximum Possible Sum
If each of \( \sin^{-1} x \), \( \sin^{-1} y \), and \( \sin^{-1} z \) is \( \frac{\pi}{2} \), then:
\[ \sin^{-1} x + \sin^{-1} y + \sin^{-1} z = \frac{\pi}{2} + \frac{\pi}{2} + \frac{\pi}{2} = \frac{3\pi}{2} \]
This means that for the equation to hold, each of \( x \), \( y \), and \( z \) must be equal to 1, since \( \sin^{-1}(1) = \frac{\pi}{2} \).

### Step 3: Conclusion
Thus, we conclude that:
\[ x = 1, \quad y = 1, \quad z = 1 \]

### Final Answer
The values of \( x \), \( y \), and \( z \) are all equal to 1.

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