If `sin^(-1) x + sin^(-1) y + sin^(-1) z = pi" , prove that " x sqrt(1-x^(2) ) + y sqrt(1 - y^(2)) + zsqrt( 1 - z^(2)) = 2 xyz`.

Answer

To prove that if \( \sin^{-1} x + \sin^{-1} y + \sin^{-1} z = \pi \), then \( x \sqrt{1 - x^2} + y \sqrt{1 - y^2} + z \sqrt{1 - z^2} = 2xyz \), we will follow these steps:

### Step 1: Set up the equations
Let:
\[ a = \sin^{-1} x, \quad b = \sin^{-1} y, \quad c = \sin^{-1} z \]
Given that:
\[ a + b + c = \pi \]

### Step 2: Use the identity for sine
From the identity for sine, when \( a + b + c = \pi \):
\[ \sin^2 a + \sin^2 b + \sin^2 c = 4 \sin a \sin b \sin c \]

### Step 3: Substitute sine values
We know:
\[ \sin a = x, \quad \sin b = y, \quad \sin c = z \]
Thus, we can rewrite the identity as:
\[ x^2 + y^2 + z^2 = 4xyz \sin a \sin b \sin c \]

### Step 4: Express sine in terms of cosine
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ \cos a = \sqrt{1 - x^2}, \quad \cos b = \sqrt{1 - y^2}, \quad \cos c = \sqrt{1 - z^2} \]

### Step 5: Substitute into the identity
Now substituting the values of sine and cosine into the identity:
\[ x \sqrt{1 - x^2} + y \sqrt{1 - y^2} + z \sqrt{1 - z^2} = 2xyz \]

### Step 6: Final expression
Thus, we have:
\[ x \sqrt{1 - x^2} + y \sqrt{1 - y^2} + z \sqrt{1 - z^2} = 2xyz \]

### Conclusion
This completes the proof.

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