In a `Delta ABC`, Median `AD _|_` side AB. Find the value of `(tan.A)/(tan.B)`.

Answer

To solve the problem, we need to find the value of \(\frac{\tan A}{\tan B}\) in triangle \(ABC\) where median \(AD\) is perpendicular to side \(AB\).

### Step-by-Step Solution:

1. **Understanding the Triangle and Median**:
- Let triangle \(ABC\) be such that \(AD\) is the median from vertex \(A\) to side \(BC\).
- Since \(AD\) is perpendicular to \(AB\), we have \(\angle ADB = 90^\circ\).

2. **Labeling Angles**:
- Let \(\angle A = A\), \(\angle B = B\), and \(\angle C = C\).
- Since \(AD\) is perpendicular to \(AB\), we can denote \(\angle ADB = 90^\circ\).

3. **Using Angle Relationships**:
- The angle at point \(D\) can be expressed in terms of angles \(A\) and \(B\):
\[ \angle ADC = 90^\circ - B \]
- The angle \(D\) can also be expressed as:
\[ \angle DAB = A - 90^\circ \]

4. **Applying the M-N Theorem**:
- According to the M-N theorem, if we consider the angles around point \(D\), we have:
\[ \tan A = \frac{m}{n} \cot \theta \]
where \(m\) and \(n\) are the segments created by the median on side \(BC\).

5. **Finding the Relationship**:
- From the properties of triangles and the relationships of angles, we can derive:
\[ \tan A = \frac{1}{\cot (90^\circ - B)} = \tan B \]
- Thus, we can express:
\[ \frac{\tan A}{\tan B} = \frac{1}{1} = 1 \]

6. **Final Result**:
- Therefore, the value of \(\frac{\tan A}{\tan B}\) is:
\[ \frac{\tan A}{\tan B} = 1 \]