If `omega` is one of the angles between the normals to the ellipse `(x^2)/(a^2)+(y^2)/(b^2)=1' at the point whose eccentric angles are `theta` and `pi/2+theta` , then prove that `(2cotomega)/(sin2theta)=(e^2)/(sqrt(1-e^2))`

Explanation

The equations of the normals to the ellipse `(x^(2))/(a^(2))+(y^(2))/(gb^(2))=1` at the points whose eccentric angles are `theta and (pi)/(2)+thea` are, respectively.
`ax sec theta-"by cosec"theta=a^(2)-b^(2)`
and `ax "cosec" theta-"by" sec theta=a^(2)-b^(2)`
Sicnce `omega` is the angle between these two normals, we have `tan omega|((a)/(b) tan theta+(a)/(b)cot theta)/(1-(a^(2))/(b^(2)))|`
`=|(ab(tan theta+cot theta))/(b^(2)-a^(2))|`
`=|(2ab)/((sin2theta)(b^(2)-a^(2)))|`
`=(2ab)/((a^(2)-b^(2))sin 2 theta)=(2a^(2)sqrt(1-e^(2)))/(a^(2)e^(2) sin2 theta)`
`:. (2cot omega)/(sin 2 theta)=(e^(2))/(sqrt(1-e^(2)))`