A hyperbola, having the transverse axis of length $2 \sin \theta$, is confocal with the ellipse $3x^{2} + 4y^{2} = 12$. Then its equation is

$\dfrac{x^{2}}{4} + \dfrac{y^{2}}{3} = 1 \Rightarrow a^{2} = 4,\ b^{2} = 3$

$c^{2} = a^{2} - b^{2} = 1 \Rightarrow \text{foci } (\pm 1, 0)$

$\text{Transverse axis } = 2\sin\theta \Rightarrow a_h = \sin\theta$

$\text{Find the hyperbola with foci } (\pm 1, 0)$

$\text{and transverse axis } 2\sin\theta$

$\dfrac{x^{2}}{4}+\dfrac{y^{2}}{3}=1,\qquad e=\dfrac{1}{2}$

$\text{Foci } F_1(-1,0),\ F_2(1,0)$

$\text{1. Confocal } \Rightarrow \text{same foci } (\pm 1,0)$

$\text{2. Hyperbola: } c^{2} = a^{2} + b^{2}$

$\text{3. Get } a \text{ from the axis, then } b, \text{ write the equation}$

$\text{Confocal} \Rightarrow \text{foci of hyperbola } (\pm 1,0)$

$\therefore\ c = 1,\qquad c^{2} = 1$

$2a = 2\sin\theta \Rightarrow a = \sin\theta$

$a^{2} = \sin^{2}\theta$

$c^{2} = a^{2} + b^{2} \Rightarrow 1 = \sin^{2}\theta + b^{2}$

$b^{2} = 1 - \sin^{2}\theta = \cos^{2}\theta$

$\dfrac{x^{2}}{a^{2}} - \dfrac{y^{2}}{b^{2}} = 1$

$\dfrac{x^{2}}{\sin^{2}\theta} - \dfrac{y^{2}}{\cos^{2}\theta} = 1$

$x^{2}\operatorname{cosec}^{2}\theta - y^{2}\sec^{2}\theta = 1$

$\Rightarrow\ \text{option (1)}$

$\boxed{\text{Answer: } x^{2}\operatorname{cosec}^{2}\theta - y^{2}\sec^{2}\theta = 1\ \text{— option (1)}}$