If any two chords be drawn through two points on the major axis of an ellipse equidistant from the center,show that $\tan\frac{\alpha}{2}\tan\frac{\beta}{2}\tan\frac{\gamma}{2}\tan\frac{\delta}{2}=1$,where $\alpha,\beta,\gamma,\delta$ are the eccentric angles of the extremities of the chords.
I tried to solve it.Let the chords cut the major axis at $(-d,0)$ and $(d,0)$.
The equation of the chords are $y=\frac{-b}{a}\cot\frac{\alpha+\beta}{2}(x+d)$ and $y=\frac{-b}{a}\cot\frac{\gamma+\delta}{2}(x-d)$
and then i stuck.What should i do to reach the answer.Please help.
The equation to the chord of the ellipse joining two points with eccentric angles $\alpha$ and $\beta$ is $\frac{x}{a}\cos \frac{\alpha+\beta}{2}+\frac{y}{b}\sin \frac{\alpha+\beta}{2}=\cos\frac{\alpha-\beta}{2}$
Similarly,The equation to the chord of the ellipse joining two points with eccentric angles $\gamma$ and $\delta$ is $\frac{x}{a}\cos \frac{\gamma+\delta}{2}+\frac{y}{b}\sin \frac{\gamma+\delta}{2}=\cos\frac{\gamma-\delta}{2}$
Since,first chord passes through $(d,0)$ and the second chord passes through $(-d,0)$.
$\frac{d}{a}\cos \frac{\alpha+\beta}{2}=\cos\frac{\alpha-\beta}{2}$ and
$\frac{-d}{a}\cos \frac{\gamma+\delta}{2}=\cos\frac{\gamma-\delta}{2}$
$\Rightarrow \frac{\cos\frac{\alpha-\beta}{2}}{\cos\frac{\alpha+\beta}{2}}=-\frac{\cos\frac{\gamma-\delta}{2}}{\cos\frac{\gamma+\delta}{2}}$
$\Rightarrow \frac{\cos\frac{\alpha}{2}\cos\frac{\beta}{2}+\sin\frac{\alpha}{2}\sin\frac{\beta}{2}}{\cos\frac{\alpha}{2}\cos\frac{\beta}{2}-\sin\frac{\alpha}{2}\sin\frac{\beta}{2}}=-\frac{\cos\frac{\gamma}{2}\cos\frac{\delta}{2}+\sin\frac{\gamma}{2}\sin\frac{\delta}{2}}{\cos\frac{\gamma}{2}\cos\frac{\delta}{2}-\sin\frac{\gamma}{2}\sin\frac{\delta}{2}}$
Now apply componendo and dividendo on both sides,we will get the answer.