`alpha & beta ` are solutions of `a cos theta+b sin theta=c(cosalpha != cos beta)&(sin alpha != sin beta)` Then `tan((alpha+beta)/2)=?`

Options

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

Answer

To solve the problem, we need to find the value of \( \tan\left(\frac{\alpha + \beta}{2}\right) \) given that \( \alpha \) and \( \beta \) are solutions of the equation \( a \cos \theta + b \sin \theta = c \).

### Step-by-Step Solution:

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

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

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

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

5. **Use the identity \( \cos^2 \theta = 1 - \sin^2 \theta \)**:
\[ a^2 (1 - \sin^2 \theta) = c^2 - 2bc \sin \theta + b^2 \sin^2 \theta \]

6. **Rearranging gives**:
\[ a^2 - a^2 \sin^2 \theta = c^2 - 2bc \sin \theta + b^2 \sin^2 \theta \]

7. **Combine like terms**:
\[ (a^2 + b^2) \sin^2 \theta - 2bc \sin \theta + (c^2 - a^2) = 0 \]

8. **Identify coefficients for the quadratic equation**:
- Coefficient of \( \sin^2 \theta \): \( a^2 + b^2 \)
- Coefficient of \( \sin \theta \): \( -2bc \)
- Constant term: \( c^2 - a^2 \)

9. **Using the sum of roots formula**:
From the quadratic equation \( Ax^2 + Bx + C = 0 \), the sum of the roots \( \alpha + \beta \) can be given by:
\[ \sin \alpha + \sin \beta = \frac{-B}{A} = \frac{2bc}{a^2 + b^2} \]

10. **Repeat the process for \( b \sin \theta = c - a \cos \theta \)**:
\[ b \sin \theta = c - a \cos \theta \]
Squaring and rearranging gives:
\[ (b^2 + a^2) \cos^2 \theta - 2ac \cos \theta + (c^2 - b^2) = 0 \]
From this, we find:
\[ \cos \alpha + \cos \beta = \frac{2ac}{a^2 + b^2} \]

11. **Using the identities for sum of angles**:
\[ \frac{\sin \alpha + \sin \beta}{\cos \alpha + \cos \beta} = \tan\left(\frac{\alpha + \beta}{2}\right) \]

12. **Substituting the values**:
\[ \tan\left(\frac{\alpha + \beta}{2}\right) = \frac{\frac{2bc}{a^2 + b^2}}{\frac{2ac}{a^2 + b^2}} = \frac{bc}{ac} = \frac{b}{a} \]

13. **Final Result**:
\[ \tan\left(\frac{\alpha + \beta}{2}\right) = \frac{b}{a} \]