If `sin^(-1)x+sin^(-1)y+sin^(-1)z=pi`, show that `x^4+y^4+z^4+4x^2y^2z^2=2(x^2y^2+y^2z^2+z^2x^2)`

Answer

To prove the equation \( x^4 + y^4 + z^4 + 4x^2y^2z^2 = 2(x^2y^2 + y^2z^2 + z^2x^2) \) given that \( \sin^{-1}x + \sin^{-1}y + \sin^{-1}z = \pi \), we can follow these steps:

### Step 1: Define the angles
Let:
\[ \sin^{-1}x = \alpha, \quad \sin^{-1}y = \beta, \quad \sin^{-1}z = \gamma \]
Thus, we have:
\[ \alpha + \beta + \gamma = \pi \]

### Step 2: Express sine and cosine
From the definitions, we know:
\[ x = \sin \alpha, \quad y = \sin \beta, \quad z = \sin \gamma \]
Using the identity for cosine, we can express:
\[ \cos(\alpha + \beta) = -\cos\gamma \]

### Step 3: Use the cosine addition formula
Using the cosine addition formula:
\[ \cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta \]
We can substitute:
\[ \cos \alpha = \sqrt{1 - x^2}, \quad \cos \beta = \sqrt{1 - y^2}, \quad \sin \alpha = x, \quad \sin \beta = y \]
Thus:
\[ \cos(\alpha + \beta) = \sqrt{1 - x^2} \sqrt{1 - y^2} - xy \]
Setting this equal to \(-\cos \gamma\):
\[ \sqrt{1 - x^2} \sqrt{1 - y^2} - xy = -\sqrt{1 - z^2} \]

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

### Step 5: Square both sides
Squaring both sides:
\[ (1 - x^2)(1 - y^2) + 1 - z^2 + 2\sqrt{(1 - x^2)(1 - y^2)(1 - z^2)} = x^2y^2 \]
This simplifies to:
\[ 1 - x^2 - y^2 + x^2y^2 + 1 - z^2 = x^2y^2 - 2\sqrt{(1 - x^2)(1 - y^2)(1 - z^2)} \]

### Step 6: Rearranging again
Rearranging gives:
\[ x^2 + y^2 - z^2 = 2xy\sqrt{1 - z^2} \]

### Step 7: Square again
Squaring again:
\[ (x^2 + y^2 - z^2)^2 = 4x^2y^2(1 - z^2) \]
Expanding both sides:
\[ x^4 + y^4 + z^4 + 2x^2y^2 - 2x^2z^2 - 2y^2z^2 = 4x^2y^2 - 4x^2y^2z^2 \]

### Step 8: Combine terms
Rearranging gives:
\[ x^4 + y^4 + z^4 + 4x^2y^2z^2 = 2(x^2y^2 + y^2z^2 + z^2x^2) \]

Thus, we have shown that:
\[ x^4 + y^4 + z^4 + 4x^2y^2z^2 = 2(x^2y^2 + y^2z^2 + z^2x^2) \]