If `3 sin alpha=5 sin beta`, then `(tan((alpha+beta)/2))/(tan ((alpha-beta)/2))=`

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem where \( 3 \sin \alpha = 5 \sin \beta \) and we need to find \( \frac{\tan\left(\frac{\alpha + \beta}{2}\right)}{\tan\left(\frac{\alpha - \beta}{2}\right)} \), we can follow these steps:

### Step 1: Express \( \frac{\sin \alpha}{\sin \beta} \)
From the given equation \( 3 \sin \alpha = 5 \sin \beta \), we can express \( \frac{\sin \alpha}{\sin \beta} \) as follows:

\[ \frac{\sin \alpha}{\sin \beta} = \frac{5}{3} \]

**Hint:** This step involves rearranging the given equation to find the ratio of the sines of the angles.

### Step 2: Use the tangent half-angle identities
We know the identities for tangent in terms of sine:

\[ \tan\left(\frac{\alpha + \beta}{2}\right) = \frac{\sin\left(\frac{\alpha + \beta}{2}\right)}{\cos\left(\frac{\alpha + \beta}{2}\right)} \]
\[ \tan\left(\frac{\alpha - \beta}{2}\right) = \frac{\sin\left(\frac{\alpha - \beta}{2}\right)}{\cos\left(\frac{\alpha - \beta}{2}\right)} \]

### Step 3: Use the sine addition and subtraction formulas
Using the sine addition and subtraction formulas, we can express the sine terms:

\[ \sin\left(\frac{\alpha + \beta}{2}\right) = \frac{1}{2} \left( \sin \alpha + \sin \beta \right) \]
\[ \sin\left(\frac{\alpha - \beta}{2}\right) = \frac{1}{2} \left( \sin \alpha - \sin \beta \right) \]

### Step 4: Substitute the sine values
Substituting the expressions for \( \sin \alpha \) and \( \sin \beta \):

Let \( \sin \beta = x \), then \( \sin \alpha = \frac{5}{3} x \).

Now substituting into the sine addition and subtraction formulas:

\[ \sin\left(\frac{\alpha + \beta}{2}\right) = \frac{1}{2} \left( \frac{5}{3} x + x \right) = \frac{1}{2} \left( \frac{8}{3} x \right) = \frac{4}{3} x \]

\[ \sin\left(\frac{\alpha - \beta}{2}\right) = \frac{1}{2} \left( \frac{5}{3} x - x \right) = \frac{1}{2} \left( \frac{2}{3} x \right) = \frac{1}{3} x \]

### Step 5: Calculate the ratio of tangents
Now we can find the ratio:

\[ \frac{\tan\left(\frac{\alpha + \beta}{2}\right)}{\tan\left(\frac{\alpha - \beta}{2}\right)} = \frac{\frac{\sin\left(\frac{\alpha + \beta}{2}\right)}{\cos\left(\frac{\alpha + \beta}{2}\right)}}{\frac{\sin\left(\frac{\alpha - \beta}{2}\right)}{\cos\left(\frac{\alpha - \beta}{2}\right)}} = \frac{\frac{4}{3} x}{\frac{1}{3} x} = \frac{4}{1} = 4 \]

### Final Answer
Thus, we have:

\[ \frac{\tan\left(\frac{\alpha + \beta}{2}\right)}{\tan\left(\frac{\alpha - \beta}{2}\right)} = 4 \]