In a `DeltaABC, P and Q` are the mid-points of `AB and AC` respectively. If O is the circumcentre of the `Delta ABC,` then the value of `(("Area of "Delta ABC)/("Area of " Delta OPQ)) cot B cot C` equal to

Answer

To solve the problem, we need to find the value of \(\frac{\text{Area of } \Delta ABC}{\text{Area of } \Delta OPQ} \cdot \cot B \cdot \cot C\).

### Step 1: Identify the areas of the triangles
1. **Area of Triangle ABC**:
The area of triangle \(ABC\) can be expressed using the formula:
\[ \text{Area}_{ABC} = \frac{1}{2} \cdot AB \cdot AC \cdot \sin C \]
We can also express this in terms of the circumradius \(R\):
\[ \text{Area}_{ABC} = \frac{abc}{4R} \]
where \(a = BC\), \(b = AC\), and \(c = AB\).

2. **Area of Triangle OPQ**:
Since \(P\) and \(Q\) are midpoints of \(AB\) and \(AC\), respectively, the segment \(PQ\) is parallel to \(BC\) and half its length:
\[ PQ = \frac{1}{2} BC = \frac{a}{2} \]
The height from \(O\) to \(PQ\) can be calculated using the circumradius \(R\) and the angles:
\[ OQ = R \cos B \]
The angle \(PQO\) is \(90^\circ - C\), thus:
\[ \text{Area}_{OPQ} = \frac{1}{2} \cdot PQ \cdot OQ \cdot \sin(PQO) = \frac{1}{2} \cdot \frac{a}{2} \cdot R \cos B \cdot \sin(90^\circ - C) = \frac{1}{2} \cdot \frac{a}{2} \cdot R \cos B \cdot \cos C \]
Simplifying this gives:
\[ \text{Area}_{OPQ} = \frac{aR}{4} \cos B \cos C \]

### Step 2: Calculate the ratio of the areas
Now we can find the ratio of the areas:
\[ \frac{\text{Area}_{ABC}}{\text{Area}_{OPQ}} = \frac{\frac{abc}{4R}}{\frac{aR}{4} \cos B \cos C} \]
This simplifies to:
\[ \frac{abc}{4R} \cdot \frac{4}{aR \cos B \cos C} = \frac{bc}{R^2 \cos B \cos C} \]

### Step 3: Multiply by \(\cot B \cot C\)
Now we multiply this ratio by \(\cot B \cot C\):
\[ \frac{bc}{R^2 \cos B \cos C} \cdot \cot B \cot C = \frac{bc}{R^2 \cos B \cos C} \cdot \frac{\cos B \cos C}{\sin B \sin C} = \frac{bc}{R^2 \sin B \sin C} \]

### Step 4: Final simplification
Using the relationship \(R = \frac{abc}{4 \text{Area}_{ABC}}\), we can substitute and simplify:
\[ \frac{bc}{\left(\frac{abc}{4 \text{Area}_{ABC}}\right)^2 \sin B \sin C} = \frac{4 \text{Area}_{ABC}^2}{a^2 \sin B \sin C} \]

### Conclusion
After evaluating the above expressions, we find that:
\[ \frac{\text{Area}_{ABC}}{\text{Area}_{OPQ}} \cdot \cot B \cdot \cot C = 4 \]

Thus, the final answer is:
\[ \boxed{4} \]