If `A + B + C = 270^(@)`, then the value of `cos 2 A + cos 2 B + cos 2 C + 4 sin A sin B sin C` is

Options

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

Answer

To solve the problem, we start with the given equation:

**Given:**
\[ A + B + C = 270^\circ \]

We need to find the value of:
\[ \cos 2A + \cos 2B + \cos 2C + 4 \sin A \sin B \sin C \]

### Step 1: Express \( B + C \)
From the given equation, we can express \( B + C \) in terms of \( A \):
\[ B + C = 270^\circ - A \]

### Step 2: Use the Cosine Addition Formula
We can use the cosine addition formula to express \( \cos 2B + \cos 2C \):
\[ \cos 2B + \cos 2C = 2 \cos\left(\frac{2B + 2C}{2}\right) \cos\left(\frac{2B - 2C}{2}\right) \]
This simplifies to:
\[ \cos 2B + \cos 2C = 2 \cos(B + C) \cos(B - C) \]

### Step 3: Substitute \( B + C \)
Substituting \( B + C = 270^\circ - A \):
\[ \cos 2B + \cos 2C = 2 \cos(270^\circ - A) \cos(B - C) \]

Using the cosine identity:
\[ \cos(270^\circ - A) = -\sin A \]
Thus:
\[ \cos 2B + \cos 2C = 2(-\sin A) \cos(B - C) = -2 \sin A \cos(B - C) \]

### Step 4: Combine Terms
Now, we substitute this back into our expression:
\[ \cos 2A + \cos 2B + \cos 2C + 4 \sin A \sin B \sin C \]
becomes:
\[ \cos 2A - 2 \sin A \cos(B - C) + 4 \sin A \sin B \sin C \]

### Step 5: Use the Cosine Double Angle Identity
Using the identity for \( \cos 2A \):
\[ \cos 2A = 1 - 2 \sin^2 A \]
So we have:
\[ 1 - 2 \sin^2 A - 2 \sin A \cos(B - C) + 4 \sin A \sin B \sin C \]

### Step 6: Simplify Further
Now, we can factor out \( 2 \sin A \):
\[ 1 - 2 \sin A (\sin A + \cos(B - C) - 2 \sin B \sin C) \]

### Step 7: Evaluate the Expression
Notice that \( \sin A + \cos(B - C) - 2 \sin B \sin C \) can be simplified further, but we can directly evaluate the expression based on the known values.

### Final Result
After going through the calculations and simplifications, we find that:
\[ \cos 2A + \cos 2B + \cos 2C + 4 \sin A \sin B \sin C = 1 \]

Thus, the final answer is:
\[ \boxed{1} \]