If `f(x) and g(x)` are functions such that `f(x + y) = f(x) g(y) + g(x) f(y),` then in `|(f(alpha),g(alpha),f(alpha+theta)),(f(beta),g(beta),f(beta+theta)), (f(lambda),g(lambda),f(lambda+theta))|` is independent of

Options

A. `alpha`
B. `beta`
C. `lambda`
D. `theta`

Answer

To solve the problem, we need to analyze the given determinant and the properties of the functions \( f(x) \) and \( g(x) \).

### Step-by-Step Solution:

1. **Understanding the Given Condition**:
We are given that:
\[ f(x + y) = f(x)g(y) + g(x)f(y) \]
This is a functional equation that relates the functions \( f \) and \( g \).

2. **Setting Up the Determinant**:
We need to evaluate the determinant:
\[ D = \begin{vmatrix} f(\alpha) & g(\alpha) & f(\alpha + \theta) \\ f(\beta) & g(\beta) & f(\beta + \theta) \\ f(\lambda) & g(\lambda) & f(\lambda + \theta) \end{vmatrix} \]

3. **Using the Functional Equation**:
We can express \( f(\alpha + \theta) \), \( f(\beta + \theta) \), and \( f(\lambda + \theta) \) using the given functional equation:
\[ f(\alpha + \theta) = f(\alpha)g(\theta) + g(\alpha)f(\theta) \]
\[ f(\beta + \theta) = f(\beta)g(\theta) + g(\beta)f(\theta) \]
\[ f(\lambda + \theta) = f(\lambda)g(\theta) + g(\lambda)f(\theta) \]

4. **Substituting into the Determinant**:
Substitute these expressions into the determinant:
\[ D = \begin{vmatrix} f(\alpha) & g(\alpha) & f(\alpha)g(\theta) + g(\alpha)f(\theta) \\ f(\beta) & g(\beta) & f(\beta)g(\theta) + g(\beta)f(\theta) \\ f(\lambda) & g(\lambda) & f(\lambda)g(\theta) + g(\lambda)f(\theta) \end{vmatrix} \]

5. **Applying Determinant Properties**:
We can apply the determinant property \( C_3 \to C_3 - g(\theta)C_1 - f(\theta)C_2 \):
\[ D = \begin{vmatrix} f(\alpha) & g(\alpha) & 0 \\ f(\beta) & g(\beta) & 0 \\ f(\lambda) & g(\lambda) & 0 \end{vmatrix} \]

6. **Evaluating the Determinant**:
Since the third column is now all zeros, the determinant \( D \) evaluates to zero:
\[ D = 0 \]

7. **Conclusion**:
Since the determinant is zero, it is independent of \( \alpha \), \( \beta \), \( \lambda \), and \( \theta \).

### Final Answer:
The value of the determinant is independent of \( \alpha, \beta, \lambda, \) and \( \theta \).