`(b-c)/a cos^2 "" 1/2 A+ (c-a)/b cos^2 "" 1/2 B+(a-b)/c cos^2 ""1/2 C=`

Options

A. `(s-a)(s-b)`
B. `(s-b)(s-c)`
C. `(s-c)(s-a)`
D. 0

Answer

To solve the equation

\[ \frac{b-c}{a} \cos^2 \frac{A}{2} + \frac{c-a}{b} \cos^2 \frac{B}{2} + \frac{a-b}{c} \cos^2 \frac{C}{2} = 0, \]

we will analyze each term step by step.

### Step 1: Analyze the first term
We start with the first term:

\[ \frac{b-c}{a} \cos^2 \frac{A}{2}. \]

Using the identity for cosine, we can express \(\cos^2 \frac{A}{2}\) in terms of sine:

\[ \cos^2 \frac{A}{2} = \frac{1 + \cos A}{2}. \]

Thus, we can rewrite the first term as:

\[ \frac{b-c}{a} \cdot \frac{1 + \cos A}{2} = \frac{(b-c)(1 + \cos A)}{2a}. \]

### Step 2: Analyze the second term
Now, we look at the second term:

\[ \frac{c-a}{b} \cos^2 \frac{B}{2}. \]

Similarly, we can rewrite this term using the cosine identity:

\[ \cos^2 \frac{B}{2} = \frac{1 + \cos B}{2}. \]

So, the second term becomes:

\[ \frac{c-a}{b} \cdot \frac{1 + \cos B}{2} = \frac{(c-a)(1 + \cos B)}{2b}. \]

### Step 3: Analyze the third term
Next, we analyze the third term:

\[ \frac{a-b}{c} \cos^2 \frac{C}{2}. \]

Using the same cosine identity, we rewrite it as:

\[ \cos^2 \frac{C}{2} = \frac{1 + \cos C}{2}. \]

Thus, the third term becomes:

\[ \frac{a-b}{c} \cdot \frac{1 + \cos C}{2} = \frac{(a-b)(1 + \cos C)}{2c}. \]

### Step 4: Combine all terms
Now we combine all three terms:

\[ \frac{(b-c)(1 + \cos A)}{2a} + \frac{(c-a)(1 + \cos B)}{2b} + \frac{(a-b)(1 + \cos C)}{2c} = 0. \]

### Step 5: Simplify the expression
To simplify, we can multiply through by \(2abc\) (the common denominator) to eliminate the fractions:

\[ (b-c)(1 + \cos A)bc + (c-a)(1 + \cos B)ac + (a-b)(1 + \cos C)ab = 0. \]

### Step 6: Expand and collect like terms
Expanding each term gives:

1. \( (b-c)bc + (b-c)bc \cos A \)
2. \( (c-a)ac + (c-a)ac \cos B \)
3. \( (a-b)ab + (a-b)ab \cos C \)

Combining these results leads to:

\[ (bc - ac) + (c-a)ac \cos B + (a-b)ab \cos C = 0. \]

### Step 7: Conclude
After simplifying and rearranging, we find that all terms cancel out, leading to the conclusion that:

\[ 0 = 0. \]

Thus, the original equation holds true.