If `y tan (A+B+C) =x tan (A+B-C) = gamma then tan 2C=`

Options

A. `(gamma(x+y))/(lamda^(2)-xy)`
B. `(lamda(x+y))/(lamda^(2)+xy)`
C. `(lamda(x-y))/(xy-lamda^(2))`
D. `(lamda(x-y))/(xy+lamda^(2))`

Answer

To solve the problem, we need to find the value of \( \tan 2C \) given the equations:

\[ y \tan(A + B + C) = x \tan(A + B - C) = \gamma \]

### Step-by-step Solution:

1. **Define Variables**:
Let \( t = \tan(A + B + C) \) and \( k = \tan(A + B - C) \). From the given equations, we can express \( t \) and \( k \) in terms of \( \gamma \):
\[ y t = \gamma \implies t = \frac{\gamma}{y} \]
\[ x k = \gamma \implies k = \frac{\gamma}{x} \]

2. **Express \( \tan 2C \)**:
We can express \( \tan 2C \) using the identity:
\[ \tan 2C = \tan((A + B + C) - (A + B - C)) \]
This simplifies to:
\[ \tan 2C = \tan(2C) \]

3. **Use the Tangent Difference Formula**:
Using the formula for the tangent of the difference of two angles:
\[ \tan(x - y) = \frac{\tan x - \tan y}{1 + \tan x \tan y} \]
We set \( x = A + B + C \) and \( y = A + B - C \):
\[ \tan 2C = \frac{t - k}{1 + tk} \]

4. **Substitute \( t \) and \( k \)**:
Substituting the values of \( t \) and \( k \):
\[ \tan 2C = \frac{\frac{\gamma}{y} - \frac{\gamma}{x}}{1 + \left(\frac{\gamma}{y}\right)\left(\frac{\gamma}{x}\right)} \]

5. **Simplify the Expression**:
The numerator becomes:
\[ \frac{\gamma}{y} - \frac{\gamma}{x} = \gamma \left(\frac{x - y}{xy}\right) \]
The denominator becomes:
\[ 1 + \frac{\gamma^2}{xy} = \frac{xy + \gamma^2}{xy} \]
Thus, we have:
\[ \tan 2C = \frac{\gamma \left(\frac{x - y}{xy}\right)}{\frac{xy + \gamma^2}{xy}} = \frac{\gamma (x - y)}{xy + \gamma^2} \]

### Final Result:
The final expression for \( \tan 2C \) is:
\[ \tan 2C = \frac{\gamma (x - y)}{xy + \gamma^2} \]