If `alpha,beta,gamma` be the angles which a line makes with the coordinates axes, then

Options

A. A. `cos(2alpha)+cos(2beta)+cos(2gamma)-1=0`
B. B. `cos(2alpha)+cos(2beta)+cos(2gamma)-2=0`
C. C. `cos(2alpha)+cos(2beta)+cos(2gamma)+1=0`
D. D. `cos(2alpha)+cos(2beta)+cos(2gamma)+2=0`

Answer

To solve the problem, we need to analyze the relationship between the angles α, β, and γ that a line makes with the coordinate axes and how they relate to the cosine of those angles.

### Step-by-Step Solution:

1. **Understanding the Angles**:
- Let α, β, and γ be the angles that a line makes with the x-axis, y-axis, and z-axis, respectively.

2. **Using the Property of Cosines**:
- A fundamental property in three-dimensional geometry states that for any line in space, the sum of the squares of the cosines of the angles it makes with the coordinate axes is equal to 1:
\[ \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1 \]

3. **Manipulating the Equation**:
- We can manipulate this equation by multiplying both sides by 2:
\[ 2(\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma) = 2 \]
- This simplifies to:
\[ 2\cos^2 \alpha + 2\cos^2 \beta + 2\cos^2 \gamma = 2 \]

4. **Rewriting Each Term**:
- We can express \(2\cos^2 \theta\) in terms of \(1 + \cos^2 \theta\):
\[ 2\cos^2 \alpha = 1 + \cos^2 \alpha \]
\[ 2\cos^2 \beta = 1 + \cos^2 \beta \]
\[ 2\cos^2 \gamma = 1 + \cos^2 \gamma \]

5. **Substituting Back**:
- Substituting these back into our equation gives:
\[ (1 + \cos^2 \alpha) + (1 + \cos^2 \beta) + (1 + \cos^2 \gamma) = 2 \]
- This simplifies to:
\[ 3 + \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 2 \]

6. **Rearranging the Equation**:
- Rearranging this equation leads to:
\[ \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 2 - 3 \]
- Thus:
\[ \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma + 1 = 0 \]

7. **Conclusion**:
- Therefore, we find that:
\[ \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma + 1 = 0 \]
- This corresponds to option C in the question.

### Final Answer:
The correct option is:
\[ \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma + 1 = 0 \]