Equation of Hyperbola Confocal with Given Ellipse

To find the equation of the hyperbola, we follow these steps:

1. Find the foci of the given ellipse:
   The given equation of the ellipse is:
   $3 x^{2} + 4 y^{2} = 12$

Dividing throughout by $12$, we write it in the standard form:
$\frac{x^{2}}{4} + \frac{y^{2}}{3} = 1$

Comparing with the standard form $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$, we have:
$a^{2} = 4, \quad b^{2} = 3$

Since $a > b$, the major axis lies along the $x$-axis. The eccentricity $e$ of the ellipse is given by:
$e = \sqrt{1 - \frac{b^{2}}{a^{2}}} = \sqrt{1 - \frac{3}{4}} = \frac{1}{2}$

Thus, the coordinates of the foci of the ellipse are:
$(\pm ae, 0) = \left(\pm 2 \cdot \frac{1}{2}, 0\right) = (\pm 1, 0)$

1. Determine the equation of the confocal hyperbola:
   Since the hyperbola is confocal with the ellipse, its foci are also located at $(\pm 1, 0)$.

Let the equation of the hyperbola with its transverse axis along the $x$-axis be:
$\frac{x^{2}}{A^{2}} - \frac{y^{2}}{B^{2}} = 1$

The length of the transverse axis is given as $2 \sin \theta$, which implies:
$2A = 2 \sin \theta \implies A = \sin \theta$

For a hyperbola, the distance of the foci from the origin is given by $c = \sqrt{A^{2} + B^{2}}$. Since the foci are at $(\pm 1, 0)$, we have:
$c = 1 \implies A^{2} + B^{2} = 1$

Substituting $A = \sin \theta$:
$\sin^{2} \theta + B^{2} = 1 \implies B^{2} = 1 - \sin^{2} \theta = \cos^{2} \theta$

1. Substitute $A^{2}$ and $B^{2}$ into the hyperbola's equation:
   $\frac{x^{2}}{\sin^{2} \theta} - \frac{y^{2}}{\cos^{2} \theta} = 1$

Using the reciprocal trigonometric identities $\frac{1}{\sin \theta} = \operatorname{cosec} \theta$ and $\frac{1}{\cos \theta} = \sec \theta$, this can be rewritten as:
$x^{2} \operatorname{cosec}^{2} \theta - y^{2} \sec^{2} \theta = 1$

Therefore, the correct option is (A).