If `alpha,beta` are the different values of `x` satisfying `acosx+bsinx=c` then `tan((alpha+beta)/2)` is

Options

A. a+b
B. a-b
C. `b//a`
D. `a//b`

Answer

To solve the problem, we need to find the value of \( \tan\left(\frac{\alpha + \beta}{2}\right) \) given the equation \( a \cos x + b \sin x = c \) with different values \( \alpha \) and \( \beta \) satisfying this equation.

### Step-by-step Solution:

1. **Start with the given equation:**
\[ a \cos x + b \sin x = c \]

2. **Rearrange the equation:**
\[ b \sin x = c - a \cos x \]

3. **Square both sides:**
\[ b^2 \sin^2 x = (c - a \cos x)^2 \]

4. **Expand the right-hand side:**
\[ b^2 \sin^2 x = c^2 - 2ac \cos x + a^2 \cos^2 x \]

5. **Substitute \( \sin^2 x = 1 - \cos^2 x \):**
\[ b^2 (1 - \cos^2 x) = c^2 - 2ac \cos x + a^2 \cos^2 x \]

6. **Rearranging gives:**
\[ b^2 - b^2 \cos^2 x = c^2 - 2ac \cos x + a^2 \cos^2 x \]

7. **Combine like terms:**
\[ (a^2 + b^2) \cos^2 x - 2ac \cos x + (c^2 - b^2) = 0 \]

8. **This is a quadratic equation in terms of \( \cos x \):**
\[ A \cos^2 x + B \cos x + C = 0 \]
where \( A = a^2 + b^2 \), \( B = -2ac \), and \( C = c^2 - b^2 \).

9. **Using the quadratic formula, the sum of the roots \( \cos \alpha + \cos \beta \) is given by:**
\[ \cos \alpha + \cos \beta = -\frac{B}{A} = \frac{2ac}{a^2 + b^2} \]

10. **Using the formula for the sum of cosines:**
\[ \cos \alpha + \cos \beta = 2 \cos\left(\frac{\alpha + \beta}{2}\right) \cos\left(\frac{\alpha - \beta}{2}\right) \]
Equating gives:
\[ 2 \cos\left(\frac{\alpha + \beta}{2}\right) \cos\left(\frac{\alpha - \beta}{2}\right) = \frac{2ac}{a^2 + b^2} \]

11. **Now, let's consider the sine terms:**
\[ a \cos x + b \sin x = c \implies a \cos x = c - b \sin x \]
Following similar steps, we can derive:
\[ \sin \alpha + \sin \beta = \frac{2bc}{a^2 + b^2} \]

12. **Using the sine sum formula:**
\[ \sin \alpha + \sin \beta = 2 \sin\left(\frac{\alpha + \beta}{2}\right) \cos\left(\frac{\alpha - \beta}{2}\right) \]
Equating gives:
\[ 2 \sin\left(\frac{\alpha + \beta}{2}\right) \cos\left(\frac{\alpha - \beta}{2}\right) = \frac{2bc}{a^2 + b^2} \]

13. **Dividing the sine equation by the cosine equation:**
\[ \frac{\sin\left(\frac{\alpha + \beta}{2}\right)}{\cos\left(\frac{\alpha + \beta}{2}\right)} = \frac{bc}{ac} = \frac{b}{a} \]

14. **Thus, we find:**
\[ \tan\left(\frac{\alpha + \beta}{2}\right) = \frac{b}{a} \]

### Conclusion:
The value of \( \tan\left(\frac{\alpha + \beta}{2}\right) \) is:
\[ \boxed{\frac{b}{a}} \]