If `2 sin alphacos beta sin gamma=sinbeta sin(alpha+gamma),then tan alpha,tan beta and gamma` are in

Answer

To solve the equation \(2 \sin \alpha \cos \beta \sin \gamma = \sin \beta \sin(\alpha + \gamma)\) and find the relationship between \(\tan \alpha\), \(\tan \beta\), and \(\tan \gamma\), we can follow these steps:

### Step 1: Expand the Right Side
We start with the equation:
\[ 2 \sin \alpha \cos \beta \sin \gamma = \sin \beta \sin(\alpha + \gamma) \]
Using the sine addition formula, we can expand \(\sin(\alpha + \gamma)\):
\[ \sin(\alpha + \gamma) = \sin \alpha \cos \gamma + \cos \alpha \sin \gamma \]
Substituting this back into the equation gives:
\[ 2 \sin \alpha \cos \beta \sin \gamma = \sin \beta (\sin \alpha \cos \gamma + \cos \alpha \sin \gamma) \]

### Step 2: Distribute \(\sin \beta\)
Distributing \(\sin \beta\) on the right side, we have:
\[ 2 \sin \alpha \cos \beta \sin \gamma = \sin \beta \sin \alpha \cos \gamma + \sin \beta \cos \alpha \sin \gamma \]

### Step 3: Rearranging the Equation
Now, we can rearrange the equation:
\[ 2 \sin \alpha \cos \beta \sin \gamma - \sin \beta \sin \alpha \cos \gamma - \sin \beta \cos \alpha \sin \gamma = 0 \]

### Step 4: Factor Out Common Terms
We can factor out \(\sin \gamma\) from the first and last terms:
\[ \sin \gamma (2 \sin \alpha \cos \beta - \sin \beta \cos \alpha) - \sin \beta \sin \alpha \cos \gamma = 0 \]

### Step 5: Divide by \(\cos \alpha \cos \beta \cos \gamma\)
To find the relationship between \(\tan\) values, we divide the entire equation by \(\cos \alpha \cos \beta \cos \gamma\):
\[ \frac{2 \sin \alpha \cos \beta \sin \gamma}{\cos \alpha \cos \beta \cos \gamma} - \frac{\sin \beta \sin \alpha \cos \gamma}{\cos \alpha \cos \beta \cos \gamma} - \frac{\sin \beta \cos \alpha \sin \gamma}{\cos \alpha \cos \beta \cos \gamma} = 0 \]
This simplifies to:
\[ 2 \tan \alpha \tan \gamma = \tan \beta \tan \alpha + \tan \beta \tan \gamma \]

### Step 6: Rearranging to Find the Relationship
Rearranging gives:
\[ 2 \tan \alpha \tan \gamma = \tan \beta (\tan \alpha + \tan \gamma) \]

### Step 7: Conclusion
From this equation, we can conclude that \(\tan \alpha\), \(\tan \beta\), and \(\tan \gamma\) are in harmonic progression (HP). This is because if \(a\), \(b\), and \(c\) are in HP, then:
\[ \frac{2}{b} = \frac{1}{a} + \frac{1}{c} \]

### Final Result
Thus, we conclude that:
\[ \tan \alpha, \tan \beta, \tan \gamma \text{ are in HP.} \]