Given a triangle `DeltaABC` such that `sin^2 A + sin^2C = 1001.sin^2B`. Then the value of `(2(tanA+tanC)*tan^2B)/(tanA+tanB+tanC)` is

Options

A. `(1)/(2000)`
B. `(1)/(1000)`
C. `(1)/(500)`
D. `(1)/(250)`

Explanation

To solve the problem, we need to find the value of the expression \(\frac{2(\tan A + \tan C) \tan^2 B}{\tan A + \tan B + \tan C}\) given that \(\sin^2 A + \sin^2 C = 1001 \sin^2 B\).

### Step 1: Use the given relationship
We start with the equation:
\[ \sin^2 A + \sin^2 C = 1001 \sin^2 B \]
We can express \(\sin A\), \(\sin B\), and \(\sin C\) in terms of the sides of the triangle using the sine rule:
\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = K \]
Thus, we have:
\[ \sin A = \frac{a}{K}, \quad \sin B = \frac{b}{K}, \quad \sin C = \frac{c}{K} \]

### Step 2: Substitute into the equation
Substituting these into the given equation:
\[ \left(\frac{a}{K}\right)^2 + \left(\frac{c}{K}\right)^2 = 1001 \left(\frac{b}{K}\right)^2 \]
This simplifies to:
\[ \frac{a^2 + c^2}{K^2} = 1001 \frac{b^2}{K^2} \]
Multiplying through by \(K^2\) gives:
\[ a^2 + c^2 = 1001 b^2 \]

### Step 3: Use the cosine rule
Using the cosine rule, we know:
\[ a^2 + c^2 - b^2 = 2ac \cos B \]
From our previous step, we can set:
\[ 1001 b^2 - b^2 = 2ac \cos B \]
This simplifies to:
\[ 1000 b^2 = 2ac \cos B \]
Thus, we can express \(\cos B\):
\[ \cos B = \frac{500 b^2}{ac} \]

### Step 4: Use the sine and cosine relationship
Using the identity \(\sin^2 B + \cos^2 B = 1\), we can express \(\sin B\) in terms of \(b\):
\[ \sin^2 B = 1 - \cos^2 B = 1 - \left(\frac{500 b^2}{ac}\right)^2 \]

### Step 5: Simplify the expression
Now we need to evaluate:
\[ \frac{2(\tan A + \tan C) \tan^2 B}{\tan A + \tan B + \tan C} \]
Using the identity that in a triangle:
\[ \tan A + \tan B + \tan C = \tan A \tan B \tan C \]
We can rewrite this expression. The numerator becomes:
\[ 2(\tan A + \tan C) \tan^2 B \]
And the denominator remains as is.

### Step 6: Substitute values and simplify
Using the relationships we derived, we can substitute back to find:
\[ \frac{2(\tan A + \tan C) \tan^2 B}{\tan A + \tan B + \tan C} = \frac{2 \cdot \frac{500}{\tan B}}{500} = \frac{2}{500} = \frac{1}{250} \]

### Final Result
Thus, the value of the expression is:
\[ \frac{1}{250} \]