If a line makes angle `alpha, beta` and `gamma` with the coordinate axes respectively, then `cos2alpha+cos 2 beta+cos 2gamma=`

Options

A. 2
B. -1
C. 1
D. 2

Explanation

To solve the problem, we need to find the value of \( \cos 2\alpha + \cos 2\beta + \cos 2\gamma \) given that a line makes angles \( \alpha, \beta, \) and \( \gamma \) with the coordinate axes.

### Step-by-Step Solution:

1. **Understanding the relationship between angles and direction cosines**:
The direction cosines of a line making angles \( \alpha, \beta, \) and \( \gamma \) with the x, y, and z axes are given by:
\[ l = \cos \alpha, \quad m = \cos \beta, \quad n = \cos \gamma \]
According to the properties of direction cosines, we have:
\[ l^2 + m^2 + n^2 = 1 \]

2. **Using the double angle identity**:
We can express \( \cos 2\alpha, \cos 2\beta, \) and \( \cos 2\gamma \) using the double angle formula:
\[ \cos 2\theta = 2\cos^2 \theta - 1 \]
Thus, we can write:
\[ \cos 2\alpha = 2\cos^2 \alpha - 1 = 2l^2 - 1 \]
\[ \cos 2\beta = 2\cos^2 \beta - 1 = 2m^2 - 1 \]
\[ \cos 2\gamma = 2\cos^2 \gamma - 1 = 2n^2 - 1 \]

3. **Adding the expressions**:
Now, we add these three expressions together:
\[ \cos 2\alpha + \cos 2\beta + \cos 2\gamma = (2l^2 - 1) + (2m^2 - 1) + (2n^2 - 1) \]
Simplifying this, we get:
\[ = 2l^2 + 2m^2 + 2n^2 - 3 \]

4. **Substituting the value of \( l^2 + m^2 + n^2 \)**:
Since we know from the properties of direction cosines that:
\[ l^2 + m^2 + n^2 = 1 \]
We can substitute this into our equation:
\[ = 2(1) - 3 = 2 - 3 = -1 \]

5. **Final result**:
Therefore, the value of \( \cos 2\alpha + \cos 2\beta + \cos 2\gamma \) is:
\[ \cos 2\alpha + \cos 2\beta + \cos 2\gamma = -1 \]

### Conclusion:
The final answer is:
\[ \cos 2\alpha + \cos 2\beta + \cos 2\gamma = -1 \]