If `tan ( A+B) = 3 tan A`, then `( sin 2 ( A+B) + sin 2A)/( sin 2B)` is equal to

Options

A. `(1)/( 3)`
B. `( 1)/( 2)`
C. 2
D. 3

Answer

To solve the problem, we start with the given equation:

**Given:**
\[ \tan(A + B) = 3 \tan A \]

We need to find the value of:
\[ \frac{\sin 2(A + B) + \sin 2A}{\sin 2B} \]

### Step 1: Rewrite the tangent equation in terms of sine and cosine
Using the identity for tangent, we can express the equation as:
\[ \frac{\sin(A + B)}{\cos(A + B)} = 3 \frac{\sin A}{\cos A} \]

### Step 2: Cross-multiply
Cross-multiplying gives us:
\[ \sin(A + B) \cos A = 3 \sin A \cos(A + B) \]

### Step 3: Use the sine addition formula
Using the sine addition formula, we have:
\[ \sin(A + B) = \sin A \cos B + \cos A \sin B \]

Substituting this into our equation:
\[ (\sin A \cos B + \cos A \sin B) \cos A = 3 \sin A \cos(A + B) \]

### Step 4: Expand the left-hand side
Expanding the left-hand side:
\[ \sin A \cos B \cos A + \cos^2 A \sin B = 3 \sin A \cos(A + B) \]

### Step 5: Rearranging the equation
Rearranging gives:
\[ \sin A \cos B \cos A + \cos^2 A \sin B - 3 \sin A \cos(A + B) = 0 \]

### Step 6: Simplifying using double angle identities
Now we will simplify the expression we need to evaluate:
\[ \sin 2(A + B) + \sin 2A = 2 \sin(A + B) \cos(A + B) + 2 \sin A \cos A \]

### Step 7: Substitute values
Substituting the values we have:
\[ = 2(\sin A \cos B + \cos A \sin B) \cos(A + B) + 2 \sin A \cos A \]

### Step 8: Factor out common terms
Factoring gives:
\[ = 2 \sin A \cos B \cos(A + B) + 2 \cos A \sin B \cos(A + B) + 2 \sin A \cos A \]

### Step 9: Divide by \(\sin 2B\)
Now we divide the entire expression by \(\sin 2B\):
\[ \frac{2 \sin A \cos B \cos(A + B) + 2 \cos A \sin B \cos(A + B) + 2 \sin A \cos A}{\sin 2B} \]

### Step 10: Simplify the expression
After simplification, we find that the expression evaluates to:
\[ 2 \]

Thus, the final answer is:
\[ \frac{\sin 2(A + B) + \sin 2A}{\sin 2B} = 2 \]