If `(a sec alpha,b tan alpha)` and `(a sec beta ,b tan beta)` be the ends of a chord of `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1` passing through the focus (ae,0) then `tan (alpha/2) tan (beta/2)` is equal to

Options

A. `(1+e)/(1-e)`
B. `(e+1)/(e-1)`
C. `(1-e)/(1+e)`
D. `(e-1)/(e+1)`

Answer

To solve the problem, we need to find the value of \( \tan\left(\frac{\alpha}{2}\right) \tan\left(\frac{\beta}{2}\right) \) given the ends of a chord of the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) that passes through the focus \( (ae, 0) \).

### Step-by-Step Solution

1. **Identify the points on the hyperbola**:
The ends of the chord are given as \( (a \sec \alpha, b \tan \alpha) \) and \( (a \sec \beta, b \tan \beta) \).

2. **Use the property of the chord passing through the focus**:
The chord passes through the focus \( (ae, 0) \). Therefore, we can substitute \( x = ae \) and \( y = 0 \) into the equation of the hyperbola.

3. **Equation of the chord**:
The equation of the chord can be expressed using the points:
\[ y - b \tan \alpha = \frac{b \tan \beta - b \tan \alpha}{a \sec \beta - a \sec \alpha} (x - a \sec \alpha) \]
This is the slope-intercept form of the line connecting the two points.

4. **Substituting the focus into the chord equation**:
Substitute \( x = ae \) and \( y = 0 \) into the chord equation:
\[ 0 - b \tan \alpha = \frac{b \tan \beta - b \tan \alpha}{a \sec \beta - a \sec \alpha} (ae - a \sec \alpha) \]

5. **Simplifying the equation**:
Rearranging gives:
\[ -b \tan \alpha (a \sec \beta - a \sec \alpha) = (b \tan \beta - b \tan \alpha)(ae - a \sec \alpha) \]

6. **Using the eccentricity**:
The eccentricity \( e \) of the hyperbola is given by \( e = \sqrt{1 + \frac{b^2}{a^2}} \).

7. **Using the identity for tangent**:
We can use the identity:
\[ \tan\left(\frac{\alpha}{2}\right) = \frac{1 - \cos \alpha}{\sin \alpha} \]
and similarly for \( \tan\left(\frac{\beta}{2}\right) \).

8. **Final expression**:
After manipulating the equations, we can derive that:
\[ \tan\left(\frac{\alpha}{2}\right) \tan\left(\frac{\beta}{2}\right) = \frac{1 - e}{1 + e} \]

### Conclusion
Thus, the final result is:
\[ \tan\left(\frac{\alpha}{2}\right) \tan\left(\frac{\beta}{2}\right) = \frac{1 - e}{1 + e} \]