If AD is the altitude on BC and AD produced meets the circumcircle of `DeltaABC` at P where DP=x. Similarly EQ=y and FR=z. If a,b,c respectively denotes sides BC, CA and AB then `a/(2x)+b/(2y)+c/(2z)` has the value equal to

Options

A. (a)`tanA+tanB+tanC`
B. (b)`cotA+cosB+cotC`
C. (c)`cosA+cosB+cosC`
D. (d)`cosecA+cosecB+cosecC`

Answer

To solve the problem, we need to analyze the given information and apply some trigonometric identities and properties of triangles and circles.

### Step-by-Step Solution:

1. **Understanding the Geometry**:
- We have triangle \( \Delta ABC \) with altitude \( AD \) from vertex \( A \) to side \( BC \).
- The line \( AD \) is extended to meet the circumcircle of triangle \( ABC \) at point \( P \), with \( DP = x \).
- Similarly, \( EQ = y \) and \( FR = z \) are defined for points \( Q \) and \( R \).

2. **Using the Power of a Point Theorem**:
- By the Power of a Point theorem, we have:
\[ AD \cdot DP = BD \cdot DC \]
- Rearranging gives:
\[ DP = \frac{BD \cdot DC}{AD} \]
- Since \( DP = x \), we can write:
\[ x = \frac{BD \cdot DC}{AD} \]

3. **Expressing \( \frac{a}{2x} \)**:
- From the relationship above, we can express \( \frac{a}{2x} \):
\[ \frac{a}{2x} = \frac{a \cdot AD}{2 \cdot BD \cdot DC} \]

4. **Applying Similar Logic for \( b \) and \( c \)**:
- Similarly, we can express \( \frac{b}{2y} \) and \( \frac{c}{2z} \):
\[ \frac{b}{2y} = \frac{b \cdot AE}{2 \cdot CE \cdot AF} \]
\[ \frac{c}{2z} = \frac{c \cdot AF}{2 \cdot BF \cdot AE} \]

5. **Combining the Expressions**:
- Now we add these three fractions:
\[ \frac{a}{2x} + \frac{b}{2y} + \frac{c}{2z} = \frac{a \cdot AD}{2 \cdot BD \cdot DC} + \frac{b \cdot AE}{2 \cdot CE \cdot AF} + \frac{c \cdot AF}{2 \cdot BF \cdot AE} \]

6. **Using Trigonometric Identities**:
- We can relate the sides of the triangle to the tangents of the angles:
\[ \frac{a}{2x} = \frac{1}{2} \tan B + \frac{1}{2} \tan C \]
\[ \frac{b}{2y} = \frac{1}{2} \tan C + \frac{1}{2} \tan A \]
\[ \frac{c}{2z} = \frac{1}{2} \tan A + \frac{1}{2} \tan B \]

7. **Final Result**:
- Adding these gives:
\[ \frac{a}{2x} + \frac{b}{2y} + \frac{c}{2z} = \frac{1}{2} (\tan A + \tan B + \tan C) \]
- Since the sum of the angles in a triangle is \( 180^\circ \), we can conclude that:
\[ \frac{a}{2x} + \frac{b}{2y} + \frac{c}{2z} = 1 \]

### Conclusion:
The value of \( \frac{a}{2x} + \frac{b}{2y} + \frac{c}{2z} \) is equal to \( 1 \).