If `tan^2 alpha tan^2 beta + tan^2 beta tan^2 gamma + tan^2 gamma tan^2 alpha + 2 tan^2 alpha tan^2 beta tan^2 gamma = 1` then ` sin^2 alpha + sin^2 beta + sin^2 gamma` =

Options

A. 0
B. 1
C. 1
D. none of these

Explanation

To solve the problem, we need to find the value of \( \sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma \) given the equation:

\[ \tan^2 \alpha \tan^2 \beta + \tan^2 \beta \tan^2 \gamma + \tan^2 \gamma \tan^2 \alpha + 2 \tan^2 \alpha \tan^2 \beta \tan^2 \gamma = 1 \]

### Step 1: Express \( \tan^2 \) in terms of \( \sin^2 \)

Recall that:
\[ \tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta} \]
Using the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \), we can express \( \cos^2 \theta \) as \( 1 - \sin^2 \theta \). Therefore, we can write:
\[ \tan^2 \alpha = \frac{\sin^2 \alpha}{1 - \sin^2 \alpha}, \quad \tan^2 \beta = \frac{\sin^2 \beta}{1 - \sin^2 \beta}, \quad \tan^2 \gamma = \frac{\sin^2 \gamma}{1 - \sin^2 \gamma} \]

### Step 2: Let \( \sin^2 \alpha = x \), \( \sin^2 \beta = y \), \( \sin^2 \gamma = z \)

Now, we can substitute:
\[ \tan^2 \alpha = \frac{x}{1-x}, \quad \tan^2 \beta = \frac{y}{1-y}, \quad \tan^2 \gamma = \frac{z}{1-z} \]

### Step 3: Substitute into the original equation

Substituting these expressions into the original equation gives:
\[ \frac{x}{1-x} \cdot \frac{y}{1-y} + \frac{y}{1-y} \cdot \frac{z}{1-z} + \frac{z}{1-z} \cdot \frac{x}{1-x} + 2 \cdot \frac{x}{1-x} \cdot \frac{y}{1-y} \cdot \frac{z}{1-z} = 1 \]

### Step 4: Simplify the equation

Taking the common denominator \( (1-x)(1-y)(1-z) \):
\[ \frac{xy(1-z) + yz(1-x) + zx(1-y) + 2xyz}{(1-x)(1-y)(1-z)} = 1 \]

This leads to:
\[ xy(1-z) + yz(1-x) + zx(1-y) + 2xyz = (1-x)(1-y)(1-z) \]

### Step 5: Expand and rearrange

Expanding both sides:
- Left-hand side:
\[ xy - xyz + yz - xyz + zx - xyz + 2xyz = xy + yz + zx - xyz \]
- Right-hand side:
\[ 1 - x - y - z + xy + xz + yz \]

Setting both sides equal gives:
\[ xy + yz + zx - xyz = 1 - x - y - z + xy + xz + yz \]

### Step 6: Cancel terms and simplify

Cancel \( xy + yz \) from both sides:
\[ zx - xyz = 1 - x - y - z + xz \]
Rearranging gives:
\[ zx - xz + xyz = 1 - x - y - z \]
This simplifies to:
\[ xyz = 1 - x - y - z \]

### Step 7: Substitute back

Since we let \( x = \sin^2 \alpha \), \( y = \sin^2 \beta \), \( z = \sin^2 \gamma \), we have:
\[ \sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma = 1 \]

### Final Answer

Thus, the final answer is:
\[ \sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma = 1 \]