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(A) The equation of the chord joining points with eccentric angles $\alpha$ and $\beta$ is $\frac{x}{a} \cos \frac{\alpha + \beta}{2} + \frac{y}{b} \s

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$\frac{e - 1}{e + 1}$

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(A) The equation of the chord joining points with eccentric angles $\alpha$ and $\beta$ is $\frac{x}{a} \cos \frac{\alpha + \beta}{2} + \frac{y}{b} \sin \frac{\alpha + \beta}{2} = \cos \frac{\alpha - \beta}{2}$.Since it is a focal chord,it passes through the focus $(ae, 0)$.Substituting the focus,we get $e \cos \frac{\alpha + \beta}{2} = \cos \frac{\alpha - \beta}{2}$,which implies $\frac{\cos \frac{\alpha - \beta}{2}}{\cos \frac{\alpha + \beta}{2}} = \frac{e}{1}$.Applying componendo and dividendo,we get $\frac{\cos \frac{\alpha - \beta}{2} - \cos \frac{\alpha + \beta}{2}}{\cos \frac{\alpha - \beta}{2} + \cos \frac{\alpha + \beta}{2}} = \frac{e - 1}{e + 1}$.Using the sum-to-product formulas,this simplifies to $\frac{2 \sin \frac{\alpha}{2} \sin \frac{\beta}{2}}{2 \cos \frac{\alpha}{2} \cos \frac{\beta}{2}} = \frac{e - 1}{e + 1}$.Thus,$	an \frac{\alpha}{2} 	an \frac{\beta}{2} = \frac{e - 1}{e + 1}$.

If $\alpha$ and $\beta$ are the eccentric angles of the endpoints of a focal chord of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,then $\tan \frac{\alpha}{2} \tan \frac{\beta}{2} = ....$

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