If `tan beta=2sin alpha sin gamma co sec(alpha+gamma)`, then `cot alpha,cot beta,cotgamma` are in

Options

A. AP
B. GP
C. HP
D. none of these.

Explanation

To solve the problem, we start with the given equation:

\[ \tan \beta = 2 \sin \alpha \sin \gamma \cos(\alpha + \gamma) \]

### Step 1: Express \(\cot \beta\)

We know that \(\tan \beta = \frac{1}{\cot \beta}\), so we can rewrite the equation as:

\[ \cot \beta = \frac{1}{2 \sin \alpha \sin \gamma \cos(\alpha + \gamma)} \]

### Step 2: Use the identity for \(\cos(\alpha + \gamma)\)

Using the trigonometric identity for cosine, we can express \(\cos(\alpha + \gamma)\) in terms of sine:

\[ \cos(\alpha + \gamma) = \frac{\sin(\alpha + \gamma)}{\sin \alpha \sin \gamma} \]

### Step 3: Substitute this into the equation for \(\cot \beta\)

Substituting this back into the equation for \(\cot \beta\):

\[ \cot \beta = \frac{1}{2 \sin \alpha \sin \gamma \cdot \frac{\sin(\alpha + \gamma)}{\sin \alpha \sin \gamma}} = \frac{\sin \alpha \sin \gamma}{2 \sin(\alpha + \gamma)} \]

### Step 4: Express \(\cot \alpha\) and \(\cot \gamma\)

We know that:

\[ \cot \alpha = \frac{\cos \alpha}{\sin \alpha} \quad \text{and} \quad \cot \gamma = \frac{\cos \gamma}{\sin \gamma} \]

### Step 5: Relate \(\cot \alpha\), \(\cot \beta\), and \(\cot \gamma\)

Now we can express the relationship between \(\cot \alpha\), \(\cot \beta\), and \(\cot \gamma\):

\[ 2 \cot \beta = \frac{\cos \gamma}{\sin \gamma} + \frac{\cos \alpha}{\sin \alpha} \]

This can be rewritten as:

\[ 2 \cot \beta = \cot \gamma + \cot \alpha \]

### Step 6: Conclude that \(\cot \alpha\), \(\cot \beta\), and \(\cot \gamma\) are in Arithmetic Progression (AP)

From the equation \(2 \cot \beta = \cot \alpha + \cot \gamma\), we can conclude that \(\cot \alpha\), \(\cot \beta\), and \(\cot \gamma\) are in Arithmetic Progression (AP).

### Final Result

Thus, we have shown that \(\cot \alpha\), \(\cot \beta\), and \(\cot \gamma\) are in AP.

---