If `tan A=2 tan B+cot B, ` then `2 tan (A-B)` is equal to

Options

A. `tan B `
B. `2 tan B `
C. `cot B `
D. `2 cos B `

Answer

To solve the problem, we need to find the value of \(2 \tan(A - B)\) given that \( \tan A = 2 \tan B + \cot B \).

### Step-by-Step Solution:

1. **Start with the given equation:**
\[ \tan A = 2 \tan B + \cot B \]

2. **Multiply both sides by \(\tan B\):**
\[ \tan A \tan B = (2 \tan B + \cot B) \tan B \]
This simplifies to:
\[ \tan A \tan B = 2 \tan^2 B + 1 \]
(since \(\cot B \tan B = 1\)).

3. **Rearranging the equation:**
\[ \tan A \tan B - 2 \tan^2 B - 1 = 0 \]
This can be considered as a quadratic equation in \(\tan B\).

4. **Using the identity for \(\tan(A - B)\):**
The formula for \(\tan(A - B)\) is:
\[ \tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \]

5. **Substituting \(\tan A\) and simplifying:**
From step 1, we have:
\[ \tan(A - B) = \frac{(2 \tan B + \cot B) - \tan B}{1 + (2 \tan B + \cot B) \tan B} \]
Simplifying the numerator:
\[ \tan(A - B) = \frac{\tan B + \cot B}{1 + (2 \tan^2 B + 1)} \]
The denominator simplifies to:
\[ 1 + 2 \tan^2 B + 1 = 2 + 2 \tan^2 B = 2(1 + \tan^2 B) \]

6. **Final expression for \(\tan(A - B)\):**
Thus,
\[ \tan(A - B) = \frac{\tan B + \cot B}{2(1 + \tan^2 B)} \]

7. **Calculating \(2 \tan(A - B)\):**
\[ 2 \tan(A - B) = \frac{2(\tan B + \cot B)}{2(1 + \tan^2 B)} = \frac{\tan B + \cot B}{1 + \tan^2 B} \]

8. **Substituting \(\cot B\) as \(\frac{1}{\tan B}\):**
\[ 2 \tan(A - B) = \frac{\tan B + \frac{1}{\tan B}}{1 + \tan^2 B} \]
This simplifies to:
\[ 2 \tan(A - B) = \frac{\frac{\tan^2 B + 1}{\tan B}}{1 + \tan^2 B} = \frac{1}{\tan B} = \cot B \]

### Conclusion:
Thus, the value of \(2 \tan(A - B)\) is:
\[ \boxed{\cot B} \]