If `x_(1)` and `x_(2)` are two distind roots of the equation `a cos x+b sinx=c`, then `tan"" (x_(1)+x_(2))/(2)` is equal to

Options

A. `(a)/(b)`
B. `(b)/(a)`
C. `(c)/(a)`
D. `(a)/(c)`

Explanation

To solve the problem, we need to find the value of \( \tan\left(\frac{x_1 + x_2}{2}\right) \) given that \( x_1 \) and \( x_2 \) are the distinct roots of the equation \( a \cos x + b \sin x = c \).

### Step-by-Step Solution:

1. **Write the equation**:
We start with the equation given in the problem:
\[ a \cos x + b \sin x = c \]

2. **Rearrange the equation**:
Move \( c \) to the right-hand side:
\[ a \cos x + b \sin x - c = 0 \]

3. **Use the half-angle formulas**:
We can express \( \cos x \) and \( \sin x \) in terms of \( \tan\left(\frac{x}{2}\right) \):
\[ \cos x = \frac{1 - \tan^2\left(\frac{x}{2}\right)}{1 + \tan^2\left(\frac{x}{2}\right)}, \quad \sin x = \frac{2 \tan\left(\frac{x}{2}\right)}{1 + \tan^2\left(\frac{x}{2}\right)} \]
Substituting these into the equation gives:
\[ a \left(\frac{1 - \tan^2\left(\frac{x}{2}\right)}{1 + \tan^2\left(\frac{x}{2}\right)}\right) + b \left(\frac{2 \tan\left(\frac{x}{2}\right)}{1 + \tan^2\left(\frac{x}{2}\right)}\right) - c = 0 \]

4. **Multiply through by the denominator**:
Multiply the entire equation by \( 1 + \tan^2\left(\frac{x}{2}\right) \) to eliminate the fraction:
\[ a(1 - \tan^2\left(\frac{x}{2}\right)) + 2b \tan\left(\frac{x}{2}\right) - c(1 + \tan^2\left(\frac{x}{2}\right)) = 0 \]

5. **Rearranging**:
Rearranging gives us:
\[ (c - a) + (2b - c) \tan\left(\frac{x}{2}\right) - a \tan^2\left(\frac{x}{2}\right) = 0 \]

6. **Form a quadratic equation**:
This can be rewritten as a standard quadratic equation in terms of \( \tan\left(\frac{x}{2}\right) \):
\[ -a \tan^2\left(\frac{x}{2}\right) + (2b - c) \tan\left(\frac{x}{2}\right) + (c - a) = 0 \]

7. **Identify coefficients**:
Here, we can identify:
- \( A = -a \)
- \( B = 2b - c \)
- \( C = c - a \)

8. **Sum and product of roots**:
From the properties of quadratic equations, the sum of the roots \( \tan\left(\frac{x_1}{2}\right) + \tan\left(\frac{x_2}{2}\right) \) is given by:
\[ \tan\left(\frac{x_1}{2}\right) + \tan\left(\frac{x_2}{2}\right) = -\frac{B}{A} = \frac{2b - c}{-a} \]
The product of the roots \( \tan\left(\frac{x_1}{2}\right) \tan\left(\frac{x_2}{2}\right) \) is:
\[ \tan\left(\frac{x_1}{2}\right) \tan\left(\frac{x_2}{2}\right) = \frac{C}{A} = \frac{c - a}{-a} \]

9. **Use the tangent addition formula**:
We can express \( \tan\left(\frac{x_1 + x_2}{2}\right) \) using the sum and product of the roots:
\[ \tan\left(\frac{x_1 + x_2}{2}\right) = \frac{\tan\left(\frac{x_1}{2}\right) + \tan\left(\frac{x_2}{2}\right)}{1 - \tan\left(\frac{x_1}{2}\right) \tan\left(\frac{x_2}{2}\right)} \]

10. **Substituting values**:
Substituting the values we found:
\[ \tan\left(\frac{x_1 + x_2}{2}\right) = \frac{\frac{2b - c}{-a}}{1 - \frac{c - a}{-a}} = \frac{\frac{2b - c}{-a}}{\frac{a - c + a}{-a}} = \frac{2b - c}{2a - c} \]

11. **Final result**:
After simplifying, we find:
\[ \tan\left(\frac{x_1 + x_2}{2}\right) = \frac{b}{a} \]

### Conclusion:
Thus, the value of \( \tan\left(\frac{x_1 + x_2}{2}\right) \) is:
\[ \boxed{\frac{b}{a}} \]