If `2 tan A=3 tan B, then (sin 2B)/(5-cos 2B) ` is equal to

Options

A. `tan A- tan B`
B. `tan (A-B)`
C. `tan (A+B)`
D. `tan (A+2B)`

Answer

To solve the problem, we need to find the value of the expression \(\frac{\sin 2B}{5 - \cos 2B}\) given that \(2 \tan A = 3 \tan B\).

### Step-by-Step Solution:

1. **Use the double angle formulas**:
We know that:
\[ \sin 2B = 2 \sin B \cos B \]
and
\[ \cos 2B = 1 - 2 \sin^2 B \]
Therefore, we can rewrite the expression as:
\[ \frac{\sin 2B}{5 - \cos 2B} = \frac{2 \sin B \cos B}{5 - (1 - 2 \sin^2 B)} \]

2. **Simplify the denominator**:
Simplifying the denominator:
\[ 5 - (1 - 2 \sin^2 B) = 5 - 1 + 2 \sin^2 B = 4 + 2 \sin^2 B \]
Thus, the expression becomes:
\[ \frac{2 \sin B \cos B}{4 + 2 \sin^2 B} \]

3. **Factor out the common term in the denominator**:
We can factor out a 2 from the denominator:
\[ \frac{2 \sin B \cos B}{2(2 + \sin^2 B)} = \frac{\sin B \cos B}{2 + \sin^2 B} \]

4. **Express \(\sin B\) in terms of \(\tan B\)**:
From the given condition \(2 \tan A = 3 \tan B\), we can express \(\tan A\) in terms of \(\tan B\):
\[ \tan A = \frac{3}{2} \tan B \]
We also know that:
\[ \sin B = \frac{\tan B \cos B}{\sqrt{1 + \tan^2 B}} = \frac{\tan B}{\sec B} = \frac{\tan B \cos B}{1} \]

5. **Substituting back**:
Now we substitute \(\tan B\) into our expression:
\[ \frac{\sin B \cos B}{2 + \sin^2 B} = \frac{\tan B \cos^2 B}{2 + \tan^2 B \cos^2 B} \]

6. **Using the identity \(\sec^2 B = 1 + \tan^2 B\)**:
We can relate \(\sin^2 B\) and \(\cos^2 B\) using the identity:
\[ \sin^2 B = \tan^2 B \cos^2 B \]
Thus, we can simplify further.

7. **Final expression**:
After substituting and simplifying, we find that:
\[ \frac{\sin B \cos B}{2 + \sin^2 B} = \frac{\tan B}{2 + \tan^2 B} \]

### Final Answer:
The expression \(\frac{\sin 2B}{5 - \cos 2B}\) simplifies to:
\[ \frac{\tan B}{2 + \tan^2 B} \]