If equation ` x^2 tan^2 theta - (2 tan theta ) x+1=0` and ` ((1)/(1+log_(b)ac))x^(2)+((1)/(a+log_(c)ab))x+((1)/(1+log_(a)bc)-1)=0` (where a,b,c , ` gt ` ) have a common root and than 2nd equation has equal roots , then number of possible value of ` theta ` in `(0,pi)` is

Answer

To solve the problem, we need to analyze the two equations given and find the possible values of \( \theta \) in the interval \( (0, \pi) \).

### Step 1: Analyze the first equation
The first equation is:
\[ x^2 \tan^2 \theta - 2 \tan \theta \cdot x + 1 = 0 \]
This is a quadratic equation in \( x \). For this equation to have a common root with the second equation, we need to determine the conditions under which it has roots.

### Step 2: Find the discriminant of the first equation
The discriminant \( D \) of a quadratic equation \( ax^2 + bx + c = 0 \) is given by:
\[ D = b^2 - 4ac \]
For our equation:
- \( a = \tan^2 \theta \)
- \( b = -2 \tan \theta \)
- \( c = 1 \)

Thus, the discriminant becomes:
\[ D = (-2 \tan \theta)^2 - 4(\tan^2 \theta)(1) = 4 \tan^2 \theta - 4 \tan^2 \theta = 0 \]
Since \( D = 0 \), this implies that the first equation has equal roots.

### Step 3: Solve for the common root
Since the first equation has equal roots, we can find the root using the formula:
\[ x = \frac{-b}{2a} = \frac{2 \tan \theta}{2 \tan^2 \theta} = \frac{1}{\tan \theta} \]
Let this common root be \( x_0 = \frac{1}{\tan \theta} \).

### Step 4: Analyze the second equation
The second equation is:
\[ \frac{1}{1 + \log_b(ac)} x^2 + \frac{1}{a + \log_c(ab)} x + \left(\frac{1}{1 + \log_a(bc)} - 1\right) = 0 \]
For this equation to have equal roots, its discriminant must also be zero.

### Step 5: Find the discriminant of the second equation
Let:
- \( A = \frac{1}{1 + \log_b(ac)} \)
- \( B = \frac{1}{a + \log_c(ab)} \)
- \( C = \frac{1}{1 + \log_a(bc)} - 1 \)

The discriminant \( D_2 \) of the second equation is:
\[ D_2 = B^2 - 4AC \]
Setting \( D_2 = 0 \) gives us the condition for equal roots.

### Step 6: Solve for \( \theta \)
From the first equation, we found that:
\[ \tan \theta = 1 \implies \theta = \frac{\pi}{4} \]
Now, we need to check if there are other possible values of \( \theta \) in the interval \( (0, \pi) \).

### Step 7: Determine the number of solutions
The function \( \tan \theta \) is equal to 1 at \( \theta = \frac{\pi}{4} \) and is periodic with a period of \( \pi \). However, within the interval \( (0, \pi) \), the only solution is:
\[ \theta = \frac{\pi}{4} \]

### Final Answer
Thus, the number of possible values of \( \theta \) in the interval \( (0, \pi) \) is:
\[ \boxed{1} \]