In a triangle ABC, a (b cos C - c cos B) =

Options

A. `a^(2)`
B. `b^(2) - c^(2)`
C. 0
D. None of these

Answer

To solve the problem \( a(b \cos C - c \cos B) \) in triangle \( ABC \), we can follow these steps:

### Step 1: Write down the expression
We start with the expression given in the problem:
\[ a(b \cos C - c \cos B) \]

### Step 2: Substitute values for \( \cos C \) and \( \cos B \)
Using the cosine rule, we know:
\[ \cos C = \frac{a^2 + b^2 - c^2}{2ab} \]
\[ \cos B = \frac{a^2 + c^2 - b^2}{2ac} \]

### Step 3: Substitute these values into the expression
Substituting these values into the expression, we get:
\[ a \left( b \left( \frac{a^2 + b^2 - c^2}{2ab} \right) - c \left( \frac{a^2 + c^2 - b^2}{2ac} \right) \right) \]

### Step 4: Simplify the expression
This simplifies to:
\[ a \left( \frac{b(a^2 + b^2 - c^2)}{2ab} - \frac{c(a^2 + c^2 - b^2)}{2ac} \right) \]
\[ = a \left( \frac{a^2 + b^2 - c^2}{2} - \frac{a^2 + c^2 - b^2}{2} \right) \]

### Step 5: Combine the fractions
Now, we can combine the fractions:
\[ = a \left( \frac{(a^2 + b^2 - c^2) - (a^2 + c^2 - b^2)}{2} \right) \]
\[ = a \left( \frac{b^2 - c^2}{2} \right) \]

### Step 6: Final simplification
Thus, we have:
\[ = \frac{a(b^2 - c^2)}{2} \]

### Conclusion
The final expression simplifies to:
\[ \frac{a(b^2 - c^2)}{2} \]

### Final Answer
The expression \( a(b \cos C - c \cos B) \) simplifies to \( \frac{a(b^2 - c^2)}{2} \).