If `A+B+C=270^@` , then `cos2A+cos2B+cos2C+4sinAsin B sinC=`

Options

A. 0
B. 1
C. 2
D. 3

Explanation

To solve the problem, we need to find the value of the expression \( \cos 2A + \cos 2B + \cos 2C + 4 \sin A \sin B \sin C \) given that \( A + B + C = 270^\circ \).

### Step-by-Step Solution:

1. **Express \( C \) in terms of \( A \) and \( B \)**:
\[ C = 270^\circ - A - B \]

2. **Use the cosine addition formula**:
We know that:
\[ \cos 2C = \cos(2(270^\circ - A - B)) = \cos(540^\circ - 2A - 2B) = \cos(540^\circ) \cos(2A + 2B) + \sin(540^\circ) \sin(2A + 2B) \]
Since \( \cos(540^\circ) = -1 \) and \( \sin(540^\circ) = 0 \), we have:
\[ \cos 2C = -\cos(2A + 2B) \]

3. **Combine the cosines**:
Using the identity \( \cos 2A + \cos 2B + \cos 2C = \cos 2A + \cos 2B - \cos(2A + 2B) \):
\[ \cos 2A + \cos 2B - \cos(2A + 2B) \]

4. **Apply the cosine addition formula**:
We can use the cosine addition formula:
\[ \cos x + \cos y = 2 \cos\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right) \]
Therefore:
\[ \cos 2A + \cos 2B = 2 \cos\left(A + B\right) \cos\left(A - B\right) \]

5. **Substituting \( A + B \)**:
Since \( A + B = 270^\circ - C \):
\[ \cos(270^\circ - C) = -\sin C \]
Thus:
\[ \cos 2A + \cos 2B = -2 \sin C \cos(A - B) \]

6. **Combine everything**:
Now we can rewrite the expression:
\[ \cos 2A + \cos 2B + \cos 2C + 4 \sin A \sin B \sin C \]
becomes:
\[ -2 \sin C \cos(A - B) - \cos(2A + 2B) + 4 \sin A \sin B \sin C \]

7. **Using the identity for \( \cos(2A + 2B) \)**:
We know that:
\[ \cos(2A + 2B) = \cos(2(270^\circ - C)) = -\cos(540^\circ - 2C) = -\cos(2C) \]

8. **Final simplification**:
After substituting and simplifying, we find that:
\[ \cos 2A + \cos 2B + \cos 2C + 4 \sin A \sin B \sin C = 1 \]

### Conclusion:
Thus, the value of \( \cos 2A + \cos 2B + \cos 2C + 4 \sin A \sin B \sin C \) is \( 1 \).