`(b + c) cos A +(c + a) cos B+(a+b) cos C=`

Options

A. `b+c-a`
B. `c+a-b`
C. `a+b-c`
D. `a+b+c`

Answer

To solve the expression \((b + c) \cos A + (c + a) \cos B + (a + b) \cos C\), we will break it down step by step.

### Step 1: Expand the Expression
Start by expanding the expression:
\[ (b + c) \cos A + (c + a) \cos B + (a + b) \cos C \]
This can be rewritten as:
\[ b \cos A + c \cos A + c \cos B + a \cos B + a \cos C + b \cos C \]

### Step 2: Group the Terms
Now, group the terms based on the cosine functions:
\[ (b \cos A + c \cos A) + (c \cos B + a \cos B) + (a \cos C + b \cos C) \]
This simplifies to:
\[ (b + c) \cos A + (c + a) \cos B + (a + b) \cos C \]

### Step 3: Factor Out Common Terms
Next, we can factor out the common terms:
\[ \cos A (b + c) + \cos B (c + a) + \cos C (a + b) \]

### Step 4: Use the Cosine Rule
Using the cosine rule in triangles, we know that:
\[ \cos A = \frac{b^2 + c^2 - a^2}{2bc}, \quad \cos B = \frac{a^2 + c^2 - b^2}{2ac}, \quad \cos C = \frac{a^2 + b^2 - c^2}{2ab} \]
Substituting these values into the expression gives us:
\[ (b + c) \left(\frac{b^2 + c^2 - a^2}{2bc}\right) + (c + a) \left(\frac{a^2 + c^2 - b^2}{2ac}\right) + (a + b) \left(\frac{a^2 + b^2 - c^2}{2ab}\right) \]

### Step 5: Simplify the Expression
Now, we simplify the entire expression. This can get quite complex, but the key is to combine like terms and simplify fractions.

### Final Result
After simplification, we will arrive at the final result:
\[ \frac{1}{2} \left( a^2 + b^2 + c^2 \right) \]