If a `cos 2 theta + b sin 2 theta =c ` has ` alpha and beta ` as its solution, then the value of `tan alpha + tan beta` is

Options

A. `(c+a)/(2b)`
B. `(2b)/( c+a)`
C. `(c-a)/(2b)`
D. `(b)/(c+a)`

Answer

To solve the equation \( a \cos 2\theta + b \sin 2\theta = c \) and find the value of \( \tan \alpha + \tan \beta \) where \( \alpha \) and \( \beta \) are the solutions, we can follow these steps:

### Step-by-Step Solution:

1. **Use Trigonometric Identities**:
We know the identities for \( \cos 2\theta \) and \( \sin 2\theta \):
\[ \cos 2\theta = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} \]
\[ \sin 2\theta = \frac{2\tan \theta}{1 + \tan^2 \theta} \]

2. **Substitute into the Given Equation**:
Substitute these identities into the equation:
\[ a \left( \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} \right) + b \left( \frac{2\tan \theta}{1 + \tan^2 \theta} \right) = c \]

3. **Combine the Terms**:
Multiply through by \( 1 + \tan^2 \theta \) to eliminate the denominator:
\[ a(1 - \tan^2 \theta) + 2b \tan \theta = c(1 + \tan^2 \theta) \]

4. **Rearrange the Equation**:
Rearranging gives:
\[ a - a \tan^2 \theta + 2b \tan \theta - c - c \tan^2 \theta = 0 \]
Combine like terms:
\[ (a - c) - (a + c) \tan^2 \theta + 2b \tan \theta = 0 \]

5. **Form a Quadratic Equation**:
This can be rewritten as:
\[ -(a + c) \tan^2 \theta + 2b \tan \theta + (a - c) = 0 \]
Let \( x = \tan \theta \). The equation becomes:
\[ -(a + c)x^2 + 2bx + (a - c) = 0 \]

6. **Identify Coefficients**:
The coefficients of the quadratic equation are:
- \( A = -(a + c) \)
- \( B = 2b \)
- \( C = a - c \)

7. **Use the Sum of Roots Formula**:
The sum of the roots \( \tan \alpha + \tan \beta \) is given by the formula:
\[ \tan \alpha + \tan \beta = -\frac{B}{A} \]
Substituting the coefficients:
\[ \tan \alpha + \tan \beta = -\frac{2b}{-(a + c)} = \frac{2b}{a + c} \]

8. **Final Result**:
Thus, the value of \( \tan \alpha + \tan \beta \) is:
\[ \tan \alpha + \tan \beta = \frac{2b}{a + c} \]