If `cos alpha, cos beta, cos gamma` are the direction cosines of a line then the value of `sin^(2) alpha + sin^(2) beta + sin^(2) gamma` is__________

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem, we need to find the value of \( \sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma \) given that \( \cos \alpha, \cos \beta, \cos \gamma \) are the direction cosines of a line.

### Step-by-step Solution:

1. **Understanding Direction Cosines**:
The direction cosines of a line are defined as the cosines of the angles that the line makes with the coordinate axes. Therefore, we have:
\[ \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1 \]

2. **Using the Identity**:
We know that \( \sin^2 \theta + \cos^2 \theta = 1 \). We can rewrite \( \cos^2 \alpha, \cos^2 \beta, \) and \( \cos^2 \gamma \) in terms of sine:
\[ \cos^2 \alpha = 1 - \sin^2 \alpha \]
\[ \cos^2 \beta = 1 - \sin^2 \beta \]
\[ \cos^2 \gamma = 1 - \sin^2 \gamma \]

3. **Substituting into the Identity**:
Substitute these expressions into the direction cosines equation:
\[ (1 - \sin^2 \alpha) + (1 - \sin^2 \beta) + (1 - \sin^2 \gamma) = 1 \]

4. **Simplifying the Equation**:
This simplifies to:
\[ 3 - (\sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma) = 1 \]

5. **Isolating the Sine Squares**:
Rearranging gives:
\[ \sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma = 3 - 1 \]
\[ \sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma = 2 \]

Thus, the value of \( \sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma \) is \( \boxed{2} \).