If `alpha = theta_1 +theta_2` and `x = theta_1 -theta_2` and `tan theta_1 = lambda tan theta_2` then `sin x: sin alpha` is equal to ______

Options

A. `(lambda+1)/(lambda-1)`
B. `(lambda-1)/(lambda+1)`
C. `1/lambda`
D. `lambda/(lambda+1)`

Answer

To solve the problem, we need to find the ratio \( \frac{\sin x}{\sin \alpha} \) given the relationships between \( \alpha \), \( x \), \( \theta_1 \), and \( \theta_2 \).

### Step-by-Step Solution:

1. **Define the Variables**:
We have:
\[ \alpha = \theta_1 + \theta_2 \]
\[ x = \theta_1 - \theta_2 \]
And we are given:
\[ \tan \theta_1 = \lambda \tan \theta_2 \]

2. **Express \( \theta_1 \) and \( \theta_2 \)**:
From the definitions of \( \alpha \) and \( x \), we can express \( \theta_1 \) and \( \theta_2 \) in terms of \( \alpha \) and \( x \):
- Adding the equations:
\[ \alpha + x = 2\theta_1 \implies \theta_1 = \frac{\alpha + x}{2} \]
- Subtracting the equations:
\[ \alpha - x = 2\theta_2 \implies \theta_2 = \frac{\alpha - x}{2} \]

3. **Substituting into the Tangent Equation**:
Substitute \( \theta_1 \) and \( \theta_2 \) into the equation \( \tan \theta_1 = \lambda \tan \theta_2 \):
\[ \tan\left(\frac{\alpha + x}{2}\right) = \lambda \tan\left(\frac{\alpha - x}{2}\right) \]

4. **Using the Tangent Addition and Subtraction Formulas**:
We can use the tangent addition and subtraction formulas:
\[ \tan\left(\frac{\alpha + x}{2}\right) = \frac{\tan\left(\frac{\alpha}{2}\right) + \tan\left(\frac{x}{2}\right)}{1 - \tan\left(\frac{\alpha}{2}\right) \tan\left(\frac{x}{2}\right)} \]
\[ \tan\left(\frac{\alpha - x}{2}\right) = \frac{\tan\left(\frac{\alpha}{2}\right) - \tan\left(\frac{x}{2}\right)}{1 + \tan\left(\frac{\alpha}{2}\right) \tan\left(\frac{x}{2}\right)} \]

5. **Setting Up the Ratio**:
We want to find \( \frac{\sin x}{\sin \alpha} \). We can use the identity:
\[ \sin x = 2 \sin\left(\frac{x}{2}\right) \cos\left(\frac{x}{2}\right) \]
\[ \sin \alpha = 2 \sin\left(\frac{\alpha}{2}\right) \cos\left(\frac{\alpha}{2}\right) \]
Therefore, the ratio becomes:
\[ \frac{\sin x}{\sin \alpha} = \frac{2 \sin\left(\frac{x}{2}\right) \cos\left(\frac{x}{2}\right)}{2 \sin\left(\frac{\alpha}{2}\right) \cos\left(\frac{\alpha}{2}\right)} = \frac{\sin\left(\frac{x}{2}\right) \cos\left(\frac{x}{2}\right)}{\sin\left(\frac{\alpha}{2}\right) \cos\left(\frac{\alpha}{2}\right)} \]

6. **Final Expression**:
Thus, the final expression for the ratio \( \frac{\sin x}{\sin \alpha} \) simplifies to:
\[ \frac{\sin x}{\sin \alpha} = \frac{\tan\left(\frac{x}{2}\right)}{\tan\left(\frac{\alpha}{2}\right)} \]

### Final Answer:
The ratio \( \frac{\sin x}{\sin \alpha} \) is equal to \( \frac{\tan\left(\frac{x}{2}\right)}{\tan\left(\frac{\alpha}{2}\right)} \).