If `msintheta=nsin(theta+2alpha)`, then prove that `tan(theta+alpha)cotalpha=`

Options

A. `(m+n)/(2m)`
B. `(m-n)/(2n)`
C. `(m+n)/(m-n)`
D. `(m-n)/(m+n)`

Explanation

To prove that \( \tan(\theta + \alpha) \cdot \cot(\alpha) = \frac{M + N}{M - N} \) given that \( M \sin \theta = N \sin(\theta + 2\alpha) \), we can follow these steps:

### Step 1: Start with the given equation
We have:
\[ M \sin \theta = N \sin(\theta + 2\alpha) \]

### Step 2: Rearranging the equation
We can express \( \frac{M}{N} \) in terms of sine functions:
\[ \frac{M}{N} = \frac{\sin(\theta + 2\alpha)}{\sin \theta} \]

### Step 3: Apply the component and dividend rule
Using the component and dividend rule, we can rewrite the fraction:
\[ \frac{M}{N} = \frac{\sin(\theta + 2\alpha) + \sin \theta}{\sin(\theta + 2\alpha) - \sin \theta} \]

### Step 4: Use the sine addition formulas
Using the sine addition formula \( \sin A + \sin B = 2 \sin\left(\frac{A + B}{2}\right) \cos\left(\frac{A - B}{2}\right) \) and \( \sin A - \sin B = 2 \cos\left(\frac{A + B}{2}\right) \sin\left(\frac{A - B}{2}\right) \):
- Let \( A = \theta + 2\alpha \) and \( B = \theta \).

We have:
\[ \sin(\theta + 2\alpha) + \sin \theta = 2 \sin\left(\theta + \alpha\right) \cos(\alpha) \]
and
\[ \sin(\theta + 2\alpha) - \sin \theta = 2 \cos\left(\theta + \alpha\right) \sin(\alpha) \]

### Step 5: Substitute back into the equation
Substituting these results back into our expression for \( \frac{M}{N} \):
\[ \frac{M}{N} = \frac{2 \sin(\theta + \alpha) \cos(\alpha)}{2 \cos(\theta + \alpha) \sin(\alpha)} \]

### Step 6: Simplify the expression
The 2's cancel out:
\[ \frac{M}{N} = \frac{\sin(\theta + \alpha) \cos(\alpha)}{\cos(\theta + \alpha) \sin(\alpha)} \]
This can be rewritten as:
\[ \frac{M}{N} = \tan(\theta + \alpha) \cdot \cot(\alpha) \]

### Step 7: Conclude the proof
Thus, we have shown that:
\[ \tan(\theta + \alpha) \cdot \cot(\alpha) = \frac{M + N}{M - N} \]
This completes the proof.

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