let `a=cosA+cosB-cos(A+B)` and `b=4sin(A/2)sin(B/2)cos((A+B)/2)` Then a-b is

Options

A. 1
B. zero
C. `-1`
D. `1/2`

Answer

To solve the problem, we need to find the value of \( a - b \) where:

\[ a = \cos A + \cos B - \cos(A + B) \]
\[ b = 4 \sin\left(\frac{A}{2}\right) \sin\left(\frac{B}{2}\right) \cos\left(\frac{A + B}{2}\right) \]

### Step 1: Simplifying \( a \)

Using the cosine addition formula, we know that:

\[ \cos A + \cos B = 2 \cos\left(\frac{A + B}{2}\right) \cos\left(\frac{A - B}{2}\right) \]

Thus, we can rewrite \( a \):

\[ a = 2 \cos\left(\frac{A + B}{2}\right) \cos\left(\frac{A - B}{2}\right) - \cos(A + B) \]

Now, we can express \( \cos(A + B) \) using the double angle formula:

\[ \cos(A + B) = 2 \cos^2\left(\frac{A + B}{2}\right) - 1 \]

Substituting this into the expression for \( a \):

\[ a = 2 \cos\left(\frac{A + B}{2}\right) \cos\left(\frac{A - B}{2}\right) - \left(2 \cos^2\left(\frac{A + B}{2}\right) - 1\right) \]

This simplifies to:

\[ a = 2 \cos\left(\frac{A + B}{2}\right) \cos\left(\frac{A - B}{2}\right) - 2 \cos^2\left(\frac{A + B}{2}\right) + 1 \]

Rearranging gives:

\[ a = 1 + 2 \cos\left(\frac{A + B}{2}\right) \cos\left(\frac{A - B}{2}\right) - 2 \cos^2\left(\frac{A + B}{2}\right) \]

### Step 2: Simplifying \( b \)

We have:

\[ b = 4 \sin\left(\frac{A}{2}\right) \sin\left(\frac{B}{2}\right) \cos\left(\frac{A + B}{2}\right) \]

### Step 3: Finding \( a - b \)

Now we need to find \( a - b \):

\[ a - b = \left(1 + 2 \cos\left(\frac{A + B}{2}\right) \cos\left(\frac{A - B}{2}\right) - 2 \cos^2\left(\frac{A + B}{2}\right)\right) - 4 \sin\left(\frac{A}{2}\right) \sin\left(\frac{B}{2}\right) \cos\left(\frac{A + B}{2}\right) \]

### Step 4: Recognizing the relationship

Notice that the term \( 4 \sin\left(\frac{A}{2}\right) \sin\left(\frac{B}{2}\right) \cos\left(\frac{A + B}{2}\right) \) can be rewritten using the product-to-sum identities. However, we can also see that \( a \) can be expressed in terms of \( b \):

\[ a = b + 1 \]

Thus, we find:

\[ a - b = 1 \]

### Final Answer

\[ \boxed{1} \]