If in a `!ABC ,a tan A + btanB =(a + b) tan((A+B)/2)` , then

Options

A. A = B
B. A = -B
C. A = 2B
D. B = 2A

Answer

To solve the problem, we start with the equation given in the triangle \( ABC \):

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

### Step 1: Rearranging the equation
We can rearrange the equation to isolate terms involving \( a \) and \( b \):

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

### Step 2: Expressing tangent in terms of sine and cosine
Using the identity \( \tan x = \frac{\sin x}{\cos x} \), we rewrite the tangents:

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

### Step 3: Finding a common denominator
To simplify the left-hand side, we can find a common denominator:

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

### Step 4: Simplifying the equation
Now we can simplify the left-hand side:

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

### Step 5: Using sine subtraction formula
We can use the sine subtraction formula:

\[ \sin(A - B) = \sin A \cos B - \cos A \sin B \]

This gives us a way to express the sine terms in a more manageable form.

### Step 6: Equating coefficients
After simplifying, we can equate the coefficients of \( a \) and \( b \) on both sides of the equation. This leads us to the conclusion that:

\[ a \sin(A - B) = b \sin(B - A) \]

### Step 7: Concluding the relationship
From the equality of the sine terms, we can conclude that:

\[ \tan A = \tan B \implies A = B \]

Thus, we find that:

\[ A = B \]

### Final Answer
The relationship between angles \( A \) and \( B \) is:

\[ A = B \]