If a line makes the angle `alpha,beta,gamma` with the axes, then what is the value of `1+cos2alpha+cos2beta+cos2gamma` equal to

Options

A. `-1`
B. 0
C. 1
D. 2

Answer

To solve the problem, we need to find the value of \(1 + \cos 2\alpha + \cos 2\beta + \cos 2\gamma\) given that a line makes angles \(\alpha\), \(\beta\), and \(\gamma\) with the x, y, and z axes respectively.

### Step-by-Step Solution:

1. **Understanding Direction Cosines**:
The direction cosines of a line making angles \(\alpha\), \(\beta\), and \(\gamma\) with the x, y, and z axes are given by:
\[ l = \cos \alpha, \quad m = \cos \beta, \quad n = \cos \gamma \]
These direction cosines satisfy the equation:
\[ \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1 \]

2. **Using the Cosine Double Angle Formula**:
We know from trigonometric identities that:
\[ \cos 2\theta = 2\cos^2 \theta - 1 \]
Therefore, we can express \(\cos^2 \alpha\), \(\cos^2 \beta\), and \(\cos^2 \gamma\) in terms of \(\cos 2\alpha\), \(\cos 2\beta\), and \(\cos 2\gamma\):
\[ \cos^2 \alpha = \frac{1 + \cos 2\alpha}{2}, \quad \cos^2 \beta = \frac{1 + \cos 2\beta}{2}, \quad \cos^2 \gamma = \frac{1 + \cos 2\gamma}{2} \]

3. **Substituting into the Direction Cosine Equation**:
Substituting these expressions into the equation for direction cosines gives:
\[ \frac{1 + \cos 2\alpha}{2} + \frac{1 + \cos 2\beta}{2} + \frac{1 + \cos 2\gamma}{2} = 1 \]
Multiplying through by 2 to eliminate the fractions:
\[ (1 + \cos 2\alpha) + (1 + \cos 2\beta) + (1 + \cos 2\gamma) = 2 \]

4. **Simplifying the Equation**:
This simplifies to:
\[ 3 + \cos 2\alpha + \cos 2\beta + \cos 2\gamma = 2 \]
Rearranging gives:
\[ \cos 2\alpha + \cos 2\beta + \cos 2\gamma = 2 - 3 = -1 \]

5. **Finding the Final Value**:
Now we can find \(1 + \cos 2\alpha + \cos 2\beta + \cos 2\gamma\):
\[ 1 + \cos 2\alpha + \cos 2\beta + \cos 2\gamma = 1 - 1 = 0 \]

### Conclusion:
Thus, the value of \(1 + \cos 2\alpha + \cos 2\beta + \cos 2\gamma\) is \(0\).