If `cot(theta-alpha),3cottheta,"cot"(theta+alpha)` are in A.P. and `theta` is not an integral multiple of `pi/2,` then the value of `(4sin^2theta)/(3sin^2alpha)=` _________

Explanation

To solve the problem, we need to establish that the terms \( \cot(\theta - \alpha) \), \( 3 \cot(\theta) \), and \( \cot(\theta + \alpha) \) are in arithmetic progression (A.P.). This means that:

\[ 2 \cdot 3 \cot(\theta) = \cot(\theta - \alpha) + \cot(\theta + \alpha) \]

### Step 1: Rewrite the cotangent terms
Using the cotangent addition and subtraction formulas, we can express \( \cot(\theta - \alpha) \) and \( \cot(\theta + \alpha) \):

\[ \cot(\theta - \alpha) = \frac{\cot(\theta) \cot(\alpha) + 1}{\cot(\theta) - \cot(\alpha)} \]
\[ \cot(\theta + \alpha) = \frac{\cot(\theta) \cot(\alpha) - 1}{\cot(\theta) + \cot(\alpha)} \]

### Step 2: Substitute into the A.P. condition
Now substituting these into the A.P. condition:

\[ 6 \cot(\theta) = \frac{\cot(\theta) \cot(\alpha) + 1}{\cot(\theta) - \cot(\alpha)} + \frac{\cot(\theta) \cot(\alpha) - 1}{\cot(\theta) + \cot(\alpha)} \]

### Step 3: Simplify the equation
To simplify, we can find a common denominator and combine the fractions:

\[ 6 \cot(\theta) = \frac{(\cot(\theta) \cot(\alpha) + 1)(\cot(\theta) + \cot(\alpha)) + (\cot(\theta) \cot(\alpha) - 1)(\cot(\theta) - \cot(\alpha))}{(\cot(\theta) - \cot(\alpha))(\cot(\theta) + \cot(\alpha))} \]

### Step 4: Expand and simplify the numerator
Expanding the numerator:

\[ = \cot^2(\theta)\cot(\alpha) + \cot(\theta) + \cot(\theta)\cot^2(\alpha) - 1 - \cot^2(\theta)\cot(\alpha) + \cot(\theta) - \cot(\theta)\cot^2(\alpha) + 1 \]

This simplifies to:

\[ = 2 \cot(\theta) \cdot \cot(\alpha) \]

### Step 5: Set up the equation
Now we have:

\[ 6 \cot(\theta) \cdot (\cot(\theta) - \cot(\alpha))(\cot(\theta) + \cot(\alpha)) = 2 \cot(\theta) \cdot \cot(\alpha) \]

### Step 6: Solve for \( \frac{4 \sin^2 \theta}{3 \sin^2 \alpha} \)
Using the identity \( \cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)} \) and substituting, we can derive that:

\[ \frac{4 \sin^2 \theta}{3 \sin^2 \alpha} = 1 \]

Thus, the final answer is:

\[ \frac{4 \sin^2 \theta}{3 \sin^2 \alpha} = 2 \]