If `sin(theta+phi)=2sin.(theta-phi),` then : `tantheta=K,tanphi,` where K=…..

Options

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

Answer

To solve the equation \( \sin(\theta + \phi) = 2 \sin(\theta - \phi) \), we will use the sine addition and subtraction formulas.

### Step 1: Apply the sine addition and subtraction formulas
Using the formulas:
\[ \sin(\theta + \phi) = \sin \theta \cos \phi + \cos \theta \sin \phi \]
\[ \sin(\theta - \phi) = \sin \theta \cos \phi - \cos \theta \sin \phi \]
we can rewrite the equation as:
\[ \sin \theta \cos \phi + \cos \theta \sin \phi = 2(\sin \theta \cos \phi - \cos \theta \sin \phi) \]

### Step 2: Expand the right side
Expanding the right side gives:
\[ \sin \theta \cos \phi + \cos \theta \sin \phi = 2 \sin \theta \cos \phi - 2 \cos \theta \sin \phi \]

### Step 3: Rearrange the equation
Now, we rearrange the equation:
\[ \sin \theta \cos \phi + \cos \theta \sin \phi + 2 \cos \theta \sin \phi = 2 \sin \theta \cos \phi \]
This simplifies to:
\[ \sin \theta \cos \phi + 3 \cos \theta \sin \phi = 2 \sin \theta \cos \phi \]

### Step 4: Move terms involving \(\sin \theta \cos \phi\) to one side
Rearranging gives:
\[ 3 \cos \theta \sin \phi = 2 \sin \theta \cos \phi - \sin \theta \cos \phi \]
This simplifies to:
\[ 3 \cos \theta \sin \phi = \sin \theta \cos \phi \]

### Step 5: Divide both sides by \(\cos \phi\) and \(\cos \theta\)
Assuming \(\cos \phi \neq 0\) and \(\cos \theta \neq 0\), we can divide both sides by \(\cos \phi \cos \theta\):
\[ \frac{3 \sin \phi}{\cos \theta} = \frac{\sin \theta}{\cos \phi} \]

### Step 6: Express in terms of tangent
This can be rewritten as:
\[ 3 \tan \phi = \tan \theta \]
Thus, we find:
\[ \tan \theta = 3 \tan \phi \]

### Conclusion
From the above steps, we have:
\[ K = 3 \]
So, \( \tan \theta = K \tan \phi \) where \( K = 3 \).

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