If β is one of the angles between the normals to the ellipse x² + 3y² = 9 at the points (3cosθ, √3sinθ) and (√3sinθ, √3cosθ); θ ∈ (0, π/2); then 2cotβsin2θ is equal to

x2+3y2=9⇒2x+6ydydx=0... Differentiating w.r.tx⇒dydx=−x3yEquation of normal is−dxdy=3yxdxdy∣∣(3cosθ,√3sinθ)=3√3sinθ𕒷cosθ=√3tanθ=m1dxdy∣∣(𕒷sinθ,√3cosθ)=3√3cosθ𕒷sinθ=−√3cotθ=m2βis the angle between the normals to the ellipse(i), thentanβ=∣∣∣m1−m21+m1m2∣∣∣=∣∣∣√3tanθ+√3cotθ1𕒷tanθcotθ∣∣∣=∣∣∣√3tanθ+√3cotθ1𕒷∣∣∣tanβ=√32|tanθ+cotθ|1cotβ=√32|tanθ+cotθ|1cotβ=√32∣∣∣sinθcosθ+cosθsinθ∣∣∣1cotβ=√32∣∣∣1sinθcosθ∣∣∣1cotβ=√3sin2θ⇒2cotβsin2θ=2√3