If `x,y,z` are respectively perpendiculars from the circumcentre on the sides of the `DeltaABC`, the value of `a/x+b/y+c/z-(abc)/(4xyz)=`

Answer

To solve the problem, we need to evaluate the expression \( \frac{a}{x} + \frac{b}{y} + \frac{c}{z} - \frac{abc}{4xyz} \), where \( x, y, z \) are the perpendiculars from the circumcenter \( O \) of triangle \( ABC \) to its sides \( BC, CA, AB \) respectively.

### Step-by-Step Solution:

1. **Understanding the Geometry**:
- Let \( O \) be the circumcenter of triangle \( ABC \).
- The perpendiculars from \( O \) to the sides \( BC, CA, AB \) are denoted as \( x, y, z \) respectively.
- The sides opposite to vertices \( A, B, C \) are denoted as \( a, b, c \).

2. **Using the Relationship of Sides and Perpendiculars**:
- The lengths of the perpendiculars can be expressed in terms of the circumradius \( R \) and the angles of the triangle:
\[ x = R \cos A, \quad y = R \cos B, \quad z = R \cos C \]

3. **Substituting into the Expression**:
- Substitute \( x, y, z \) into the expression:
\[ \frac{a}{x} = \frac{a}{R \cos A}, \quad \frac{b}{y} = \frac{b}{R \cos B}, \quad \frac{c}{z} = \frac{c}{R \cos C} \]
- Therefore, we have:
\[ \frac{a}{x} + \frac{b}{y} + \frac{c}{z} = \frac{a}{R \cos A} + \frac{b}{R \cos B} + \frac{c}{R \cos C} \]

4. **Combining the Terms**:
- Factor out \( \frac{1}{R} \):
\[ = \frac{1}{R} \left( \frac{a}{\cos A} + \frac{b}{\cos B} + \frac{c}{\cos C} \right) \]

5. **Evaluating the Second Term**:
- Now, consider the term \( \frac{abc}{4xyz} \):
\[ 4xyz = 4(R \cos A)(R \cos B)(R \cos C) = 4R^3 \cos A \cos B \cos C \]
- Thus:
\[ \frac{abc}{4xyz} = \frac{abc}{4R^3 \cos A \cos B \cos C} \]

6. **Putting Everything Together**:
- Now, substituting back into the original expression:
\[ \frac{1}{R} \left( \frac{a}{\cos A} + \frac{b}{\cos B} + \frac{c}{\cos C} \right) - \frac{abc}{4R^3 \cos A \cos B \cos C} \]

7. **Using Known Trigonometric Identities**:
- We know from trigonometric identities that:
\[ \frac{a}{\cos A} + \frac{b}{\cos B} + \frac{c}{\cos C} = 2R \left( \tan A + \tan B + \tan C \right) \]
- Thus, we can rewrite the expression as:
\[ 2 \left( \tan A + \tan B + \tan C \right) - \frac{abc}{4R^3 \cos A \cos B \cos C} \]

8. **Final Evaluation**:
- It is known that in any triangle, the sum of the tangents of the angles is equal to the product of the tangents minus the sum of the tangents:
\[ \tan A + \tan B + \tan C = \tan A \tan B \tan C \]
- Therefore, the entire expression simplifies to zero:
\[ \frac{a}{x} + \frac{b}{y} + \frac{c}{z} - \frac{abc}{4xyz} = 0 \]

### Final Answer:
\[ \frac{a}{x} + \frac{b}{y} + \frac{c}{z} - \frac{abc}{4xyz} = 0 \]