If `m sin theta = n sin (theta +2 alpha)` then `tan (theta + alpha)` is

Options

A. `(m+n)/(m-n) tan alpha`
B. `(m+n)/(m-n) tan theta`
C. `(m+n)/(m-n) cot alpha`
D. `(m+n)/(m-n) cot theta`

Answer

To solve the equation \( m \sin \theta = n \sin(\theta + 2\alpha) \) and find \( \tan(\theta + \alpha) \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ m \sin \theta = n \sin(\theta + 2\alpha) \]

### Step 2: Use the sine addition formula
Using the sine addition formula, we can expand \( \sin(\theta + 2\alpha) \):
\[ \sin(\theta + 2\alpha) = \sin \theta \cos(2\alpha) + \cos \theta \sin(2\alpha) \]
Substituting this into the equation gives:
\[ m \sin \theta = n (\sin \theta \cos(2\alpha) + \cos \theta \sin(2\alpha)) \]

### Step 3: Rearrange the equation
Rearranging the equation, we have:
\[ m \sin \theta = n \sin \theta \cos(2\alpha) + n \cos \theta \sin(2\alpha) \]
This can be rearranged to:
\[ m \sin \theta - n \sin \theta \cos(2\alpha) = n \cos \theta \sin(2\alpha) \]
Factoring out \( \sin \theta \) on the left side:
\[ \sin \theta (m - n \cos(2\alpha)) = n \cos \theta \sin(2\alpha) \]

### Step 4: Divide both sides
Assuming \( m - n \cos(2\alpha) \neq 0 \), we can divide both sides by \( \sin \theta \):
\[ m - n \cos(2\alpha) = n \frac{\cos \theta \sin(2\alpha)}{\sin \theta} \]
This simplifies to:
\[ m - n \cos(2\alpha) = n \cot \theta \sin(2\alpha) \]

### Step 5: Express \( \tan(\theta + \alpha) \)
Now, we need to find \( \tan(\theta + \alpha) \):
\[ \tan(\theta + \alpha) = \frac{\tan \theta + \tan \alpha}{1 - \tan \theta \tan \alpha} \]
Let \( \tan \theta = t \), then:
\[ \tan(\theta + \alpha) = \frac{t + \tan \alpha}{1 - t \tan \alpha} \]

### Step 6: Substitute back to find \( t \)
From our earlier equation, we can express \( t \) in terms of \( m \), \( n \), and \( \alpha \):
\[ t = \frac{m - n \cos(2\alpha)}{n \sin(2\alpha)} \]

### Step 7: Final expression for \( \tan(\theta + \alpha) \)
Substituting \( t \) back into the expression for \( \tan(\theta + \alpha) \) gives:
\[ \tan(\theta + \alpha) = \frac{\frac{m - n \cos(2\alpha)}{n \sin(2\alpha)} + \tan \alpha}{1 - \frac{m - n \cos(2\alpha)}{n \sin(2\alpha)} \tan \alpha} \]

### Conclusion
After simplifying, we find that:
\[ \tan(\theta + \alpha) = \frac{m + n}{m - n} \tan \alpha \]