If any two chords be drawn through two points on major axis of an ellipse equidistant from centre, then `tan((alpha)/(2))tan((beta)/(2))tan((gamma)/(2))tan((delta)/(2))` = _____,
( where `alpha, beta, gamma, delta` are ecentric angles of extremities of chords )

Answer

To solve the problem, we need to find the value of the expression \( \tan\left(\frac{\alpha}{2}\right) \tan\left(\frac{\beta}{2}\right) \tan\left(\frac{\gamma}{2}\right) \tan\left(\frac{\delta}{2}\right) \), where \( \alpha, \beta, \gamma, \delta \) are the eccentric angles of the extremities of two chords drawn through two points on the major axis of an ellipse that are equidistant from the center.

### Step-by-Step Solution:

1. **Identify Points on the Major Axis**:
Let the two points on the major axis of the ellipse be \( (d, 0) \) and \( (-d, 0) \), where \( d \) is the distance from the center of the ellipse.

2. **Equation of the Chords**:
The equation of the chord of the ellipse corresponding to the angles \( \alpha \) and \( \beta \) can be expressed as:
\[ \frac{x}{a} \cos\left(\frac{\alpha + \beta}{2}\right) + \frac{y}{b} \sin\left(\frac{\alpha + \beta}{2}\right) = \cos\left(\frac{\alpha - \beta}{2}\right) \]
This is Equation (1).

Similarly, for the angles \( \gamma \) and \( \delta \):
\[ \frac{x}{a} \cos\left(\frac{\gamma + \delta}{2}\right) + \frac{y}{b} \sin\left(\frac{\gamma + \delta}{2}\right) = \cos\left(\frac{\gamma - \delta}{2}\right) \]
This is Equation (2).

3. **Substituting Points into the Equations**:
Substitute the point \( (d, 0) \) into Equation (1):
\[ \frac{d}{a} \cos\left(\frac{\alpha + \beta}{2}\right) = \cos\left(\frac{\alpha - \beta}{2}\right) \]
Rearranging gives:
\[ \frac{d}{a} = \frac{\cos\left(\frac{\alpha - \beta}{2}\right)}{\cos\left(\frac{\alpha + \beta}{2}\right)} \]

Now substitute the point \( (-d, 0) \) into Equation (2):
\[ \frac{-d}{a} \cos\left(\frac{\gamma + \delta}{2}\right) = \cos\left(\frac{\gamma - \delta}{2}\right) \]
Rearranging gives:
\[ \frac{d}{a} = -\frac{\cos\left(\frac{\gamma - \delta}{2}\right)}{\cos\left(\frac{\gamma + \delta}{2}\right)} \]

4. **Equating the Two Expressions**:
Since both expressions equal \( \frac{d}{a} \), we can set them equal to each other:
\[ \frac{\cos\left(\frac{\alpha - \beta}{2}\right)}{\cos\left(\frac{\alpha + \beta}{2}\right)} = -\frac{\cos\left(\frac{\gamma - \delta}{2}\right)}{\cos\left(\frac{\gamma + \delta}{2}\right)} \]

5. **Reciprocal and Simplification**:
Taking the reciprocal of both sides gives:
\[ \frac{\cos\left(\frac{\alpha + \beta}{2}\right)}{\cos\left(\frac{\alpha - \beta}{2}\right)} = -\frac{\cos\left(\frac{\gamma + \delta}{2}\right)}{\cos\left(\frac{\gamma - \delta}{2}\right)} \]

6. **Using Trigonometric Identities**:
Using the cosine addition and subtraction formulas, we can express the cosines in terms of sines and cosines of half angles. After simplification, we find that:
\[ \tan\left(\frac{\alpha}{2}\right) \tan\left(\frac{\beta}{2}\right) \tan\left(\frac{\gamma}{2}\right) \tan\left(\frac{\delta}{2}\right) = 1 \]

### Final Answer:
Thus, the value of \( \tan\left(\frac{\alpha}{2}\right) \tan\left(\frac{\beta}{2}\right) \tan\left(\frac{\gamma}{2}\right) \tan\left(\frac{\delta}{2}\right) \) is \( \boxed{1} \).