If the eccentricity of the hyperbola x^ 2 (1 + sin θ) ^ 2 - y^ 2 ^...

$e^{2} = 1 + \dfrac{\cos^{2} \theta}{\left(1 + sin \theta\right)2} = \dfrac{4}{3} ⇒ \dfrac{\cos^{2} \theta}{\left(1 + sin \theta\right)2} = \dfrac{1}{3}$
$⇒ \left[\dfrac{cos \theta}{1 + sin \theta}\right]2 = \dfrac{1}{3} ⇒ \left[\dfrac{\sin \left(\dfrac{\pi}{2} - \theta\right)}{1 + \cos \left(\dfrac{\pi}{2} - \theta\right)}\right]2 = \dfrac{1}{3}$
$⇒ \tan^{2} \left(\dfrac{\pi}{4} - \dfrac{\theta}{2}\right) = \dfrac{1}{3}$
$\dfrac{\pi}{4} - \dfrac{θ}{2} = \dfrac{- \pi}{6}, \dfrac{\pi}{6} ⇒ \dfrac{θ}{2} = \dfrac{\pi}{4} - \dfrac{\pi}{6}, \dfrac{\pi}{4} + \dfrac{\pi}{6}$
$\theta = \dfrac{\pi}{2} \pm \dfrac{\pi}{3} = \dfrac{\pi}{6}, \dfrac{5\pi}{6}$

Hence, the required sum $= \pi$