Given \(x^2 + 3y^2 = 9\), i.e., \(\dfrac{x^2}{9} + \dfrac{y^2}{3} = 1\). The equation of tangent at point \((3\cos\theta, \sqrt{3}\sin\theta)\) is \(\dfrac{x\cos\theta}{3} + \dfrac{y\sin\theta}{\sqrt{3}} = 1\). The equation of tangent at point \((-3\sin\theta, \sqrt{3}\cos\theta)\) is \(\dfrac{-x\sin\theta}{3} + \dfrac{y\cos\theta}{\sqrt{3}} = 1\). For the two tangents to be perpendicular, which of the following is correct?