If `sin^(-1)x+sin^(-1)y+sin^(-1)z=pi` then the value of `xsqrt(1-x^(2))+ysqrt(1-y^(2))+zsqrt(1-z^(2))`

Options

A. 2xyz
B. `x^(2)+y^(2)+z^(2)`
C. `xy+yz+zx`
D. none of these

Explanation

To solve the problem, we start with the given equation:

\[ \sin^{-1} x + \sin^{-1} y + \sin^{-1} z = \pi \]

### Step 1: Define Variables
Let:
- \( a = \sin^{-1} x \)
- \( b = \sin^{-1} y \)
- \( c = \sin^{-1} z \)

From the equation, we can rewrite it as:

\[ a + b + c = \pi \]

### Step 2: Use Sine Identity
Using the sine identity, we know that:

\[ \sin(a + b + c) = \sin \pi = 0 \]

Using the sine addition formula, we have:

\[ \sin(a + b + c) = \sin a \cos(b + c) + \cos a \sin(b + c) \]

Since \( b + c = \pi - a \), we can rewrite \( \sin(b + c) \) as:

\[ \sin(b + c) = \sin(\pi - a) = \sin a \]

Thus, we can express the equation as:

\[ \sin a \cos(b + c) + \cos a \sin a = 0 \]

### Step 3: Simplify the Equation
This simplifies to:

\[ \sin a (\cos(b + c) + 1) = 0 \]

Since \( \sin a \neq 0 \) (as \( a \) is not \( 0 \) or \( \pi \)), we have:

\[ \cos(b + c) + 1 = 0 \implies \cos(b + c) = -1 \]

This implies that \( b + c = \pi \).

### Step 4: Find Values of Sine and Cosine
From the definitions:
- \( x = \sin a \)
- \( y = \sin b \)
- \( z = \sin c \)

Now, we also have:

\[ \cos a = \sqrt{1 - x^2}, \quad \cos b = \sqrt{1 - y^2}, \quad \cos c = \sqrt{1 - z^2} \]

### Step 5: Evaluate the Expression
We need to find the value of:

\[ x \sqrt{1 - x^2} + y \sqrt{1 - y^2} + z \sqrt{1 - z^2} \]

Using the identity for sine and cosine, we can express this as:

\[ x \sqrt{1 - x^2} + y \sqrt{1 - y^2} + z \sqrt{1 - z^2 = \sin a \cos a + \sin b \cos b + \sin c \cos c \]

### Step 6: Use the Identity
From the identity:

\[ \sin 2a + \sin 2b + \sin 2c = 4 \sin a \sin b \sin c \]

We can express:

\[ \sin 2a = 2 \sin a \cos a \]

Thus, we have:

\[ 2(\sin a \cos a + \sin b \cos b + \sin c \cos c) = 4 \sin a \sin b \sin c \]

### Step 7: Final Result
This means:

\[ \sin a \cos a + \sin b \cos b + \sin c \cos c = 2 \sin a \sin b \sin c \]

Substituting back, we get:

\[ x \sqrt{1 - x^2} + y \sqrt{1 - y^2} + z \sqrt{1 - z^2 = 2xyz \]

Thus, the final answer is:

\[ \boxed{2xyz} \]