If in `DeltaABC, a tan A+b tan B=(a+b) tan"1/2(A+B)`, then

Options

A. `A=B`
B. `B=C`
C. `C=A`
D. `A=B=C`

Answer

To solve the equation \( a \tan A + b \tan B = (a + b) \tan \frac{1}{2}(A + B) \) in triangle \( \Delta ABC \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ a \tan A + b \tan B = (a + b) \tan \frac{1}{2}(A + B) \]

### Step 2: Factor out common terms
We can factor out \( \tan \frac{1}{2}(A + B) \) from the right-hand side:
\[ a \tan A + b \tan B = (a + b) \tan \frac{1}{2}(A + B) \]

### Step 3: Express tangent in terms of sine and cosine
Using the identity \( \tan \theta = \frac{\sin \theta}{\cos \theta} \), we can rewrite the tangents:
\[ a \frac{\sin A}{\cos A} + b \frac{\sin B}{\cos B} = (a + b) \frac{\sin \frac{1}{2}(A + B)}{\cos \frac{1}{2}(A + B)} \]

### Step 4: Cross-multiply to eliminate the fractions
Cross-multiplying gives us:
\[ a \sin A \cos \frac{1}{2}(A + B) + b \sin B \cos \frac{1}{2}(A + B) = (a + b) \sin \frac{1}{2}(A + B) \cos A \cos B \]

### Step 5: Use the sine addition formula
Using the sine addition formula, we can express \( \sin \frac{1}{2}(A + B) \) in terms of \( \sin A \) and \( \sin B \):
\[ \sin \frac{1}{2}(A + B) = \sin \left( \frac{A + B}{2} \right) \]

### Step 6: Apply the sine rule
From the sine rule, we know that:
\[ \frac{a}{\sin A} = \frac{b}{\sin B} \]
This implies that \( a \sin B = b \sin A \).

### Step 7: Substitute back into the equation
Substituting \( a \sin B \) for \( b \sin A \) in our equation allows us to simplify further.

### Step 8: Conclude with the relationship between angles
After simplification, we find that:
\[ \tan A = \tan B \]
This implies that:
\[ A = B \]

### Final Result
Thus, we conclude that if \( a \tan A + b \tan B = (a + b) \tan \frac{1}{2}(A + B) \), then:
\[ A = B \]