If `(a-b)sin (theta+phi)=(a+b)sin(theta-phi)` then

Explanation

To solve the equation \((a-b)\sin(\theta+\phi) = (a+b)\sin(\theta-\phi)\), we will follow these steps:

### Step 1: Expand the Sine Functions
Using the sine addition and subtraction formulas:
\[ \sin(\theta + \phi) = \sin\theta \cos\phi + \cos\theta \sin\phi \]
\[ \sin(\theta - \phi) = \sin\theta \cos\phi - \cos\theta \sin\phi \]
Substituting these into the equation gives:
\[ (a-b)(\sin\theta \cos\phi + \cos\theta \sin\phi) = (a+b)(\sin\theta \cos\phi - \cos\theta \sin\phi) \]

### Step 2: Distribute Both Sides
Distributing both sides results in:
\[ (a-b)\sin\theta \cos\phi + (a-b)\cos\theta \sin\phi = (a+b)\sin\theta \cos\phi - (a+b)\cos\theta \sin\phi \]

### Step 3: Rearrange the Equation
Rearranging the equation to group similar terms gives:
\[ (a-b)\sin\theta \cos\phi - (a+b)\sin\theta \cos\phi = -(a+b)\cos\theta \sin\phi - (a-b)\cos\theta \sin\phi \]
This simplifies to:
\[ [(a-b) - (a+b)]\sin\theta \cos\phi = -[(a+b) + (a-b)]\cos\theta \sin\phi \]

### Step 4: Simplify the Coefficients
This further simplifies to:
\[ (-2b)\sin\theta \cos\phi = -2a\cos\theta \sin\phi \]
Dividing both sides by -2 gives:
\[ b\sin\theta \cos\phi = a\cos\theta \sin\phi \]

### Step 5: Rearranging to Find the Relationship
Rearranging gives:
\[ \frac{\sin\phi}{\cos\phi} = \frac{b}{a} \cdot \frac{\sin\theta}{\cos\theta} \]
This implies:
\[ \tan\phi = \frac{b}{a} \tan\theta \]

### Final Result
Thus, we have:
\[ b \tan\theta = a \tan\phi \]