If β is one of the angles between the normals to the ellipse x 2 + 3 y 2 = 9 at the points 3 cos θ , 3 sin θ and - 3 sin θ , 3 cos θ ; θ ∈ 0 , π 2 ; then 2 cot β sin 2 θ is equal to :

Step-by-step solution

Given equation of ellipse is x 2 9 + y 2 3 = 1 Normal at 3 cos θ , 3 sin θ is 3 sec θ . x - 3 cosec θ y = 6 ⇒ Slope m 1 = 3 sec θ 3 cosec θ = 3 tan θ Normal at − 3 sinθ , 3 cos θ - 3 cosec θ . x - 3 sec θ . y = 6 ⇒ Slope m 2 = - 3 cosec θ 3 sec θ = - 3 cot θ Angle between normal is β . ⇒ tan β = m 1 - m 2 1 + m 1 m 2 = 3 tan θ + 3 cot θ 1 - 3 = 3 2 sin θ cos θ ⇒ tan β = 3 sin 2 &#95