If `cos^(-1)x+cos^(-1)y+cos^(-1)z=pi`, then

Options

A. `x^(2)+y^(2)=z^(2)`
B. `x^(2)+y^(2)+z^(2)=0`
C. `x^(2)+y^(2)+z^(2)=1-2xyz`
D. None of these

Explanation

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

### Step 1: Rearranging the Equation
We start with the given equation:
\[ \cos^{-1} x + \cos^{-1} y + \cos^{-1} z = \pi \]
We can rearrange this to isolate one of the inverse cosine terms:
\[ \cos^{-1} x + \cos^{-1} y = \pi - \cos^{-1} z \]

### Step 2: Applying the Cosine Function
Next, we take the cosine of both sides:
\[ \cos(\cos^{-1} x + \cos^{-1} y) = \cos(\pi - \cos^{-1} z) \]
Using the identity \( \cos(\pi - \theta) = -\cos(\theta) \), we have:
\[ \cos(\cos^{-1} x + \cos^{-1} y) = -z \]

### Step 3: Using the Cosine Addition Formula
We can use the cosine addition formula:
\[ \cos(A + B) = \cos A \cos B - \sin A \sin B \]
Here, \( A = \cos^{-1} x \) and \( B = \cos^{-1} y \):
\[ \cos(\cos^{-1} x) \cos(\cos^{-1} y) - \sin(\cos^{-1} x) \sin(\cos^{-1} y) = -z \]
This simplifies to:
\[ x y - \sqrt{1 - x^2} \sqrt{1 - y^2} = -z \]

### Step 4: Rearranging the Equation
Rearranging gives us:
\[ xy + z = \sqrt{1 - x^2} \sqrt{1 - y^2} \]

### Step 5: Squaring Both Sides
Now, we square both sides to eliminate the square roots:
\[ (xy + z)^2 = (1 - x^2)(1 - y^2) \]
Expanding both sides:
\[ x^2y^2 + 2xyz + z^2 = 1 - x^2 - y^2 + x^2y^2 \]
We can cancel \( x^2y^2 \) from both sides:
\[ 2xyz + z^2 = 1 - x^2 - y^2 \]

### Step 6: Rearranging to Find a Relationship
Rearranging gives us:
\[ x^2 + y^2 + z^2 + 2xyz = 1 \]

### Conclusion
Thus, we have derived the relationship:
\[ x^2 + y^2 + z^2 + 2xyz = 1 \]