If in a `triangleABC, BC=(e)/(|e|)-(f)/(|f|) and AC=(2e)/(|e|): |e|ne|f|,` then the value of `cos2A+cos2B+cos2C` must be

Options

A. (a)`-1`
B. (b)`0`
C. (c)`2`
D. (d)`(-3)/(2)`

Answer

To solve the problem, we will analyze the given vectors and apply properties of triangles and trigonometric identities.

### Step-by-Step Solution

1. **Understanding the Given Vectors**:
We are given two vectors:
- \( BC = \frac{e}{|e|} - \frac{f}{|f|} \)
- \( AC = \frac{2e}{|e|} \)

2. **Using the Triangle Law of Vectors**:
In triangle \( ABC \), we can express the relationship between the sides using vector addition:
\[ AB + BC = AC \]
Rearranging gives:
\[ AB = AC - BC \]

3. **Substituting the Given Values**:
Substitute the expressions for \( AC \) and \( BC \) into the equation:
\[ AB = \frac{2e}{|e|} - \left( \frac{e}{|e|} - \frac{f}{|f|} \right) \]
Simplifying this, we have:
\[ AB = \frac{2e}{|e|} - \frac{e}{|e|} + \frac{f}{|f|} = \frac{e}{|e|} + \frac{f}{|f|} \]

4. **Finding the Dot Product**:
Now, we will find the dot product of vectors \( AB \) and \( BC \):
\[ AB \cdot BC = \left( \frac{e}{|e|} + \frac{f}{|f|} \right) \cdot \left( \frac{e}{|e|} - \frac{f}{|f|} \right) \]
Expanding this using the distributive property:
\[ AB \cdot BC = \frac{e}{|e|} \cdot \frac{e}{|e|} - \frac{e}{|e|} \cdot \frac{f}{|f|} + \frac{f}{|f|} \cdot \frac{e}{|e|} - \frac{f}{|f|} \cdot \frac{f}{|f|} \]
This simplifies to:
\[ AB \cdot BC = 1 - 0 - 0 - 1 = 0 \]
Thus, \( AB \) is perpendicular to \( BC \).

5. **Determining Angles**:
Since \( AB \) is perpendicular to \( BC \), we have \( \angle B = 90^\circ \).

6. **Using the Cosine of Angles**:
We know that:
\[ \cos 2A + \cos 2B + \cos 2C = 1 + 1 + \cos 2C = 0 \]
Since \( \cos 2B = \cos 180^\circ = -1 \).

7. **Final Calculation**:
Therefore, we have:
\[ \cos 2A + (-1) + \cos 2C = 0 \]
Rearranging gives:
\[ \cos 2A + \cos 2C = 1 \]

8. **Conclusion**:
Since we have established that \( \angle B = 90^\circ \), we conclude that:
\[ \cos 2A + \cos 2B + \cos 2C = -1 \]

### Final Answer:
Thus, the value of \( \cos 2A + \cos 2B + \cos 2C \) must be \( -1 \).

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