Step-by-step Solution:

1. Analyze the given ellipse equation. The equation of the ellipse is given as $3{x^2} + 4{y^2} = 12$. To find its properties, we first write it in the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. Dividing the entire equation by 12, we get: $\frac{3x^2}{12} + \frac{4y^2}{12} = 1$ $\frac{x^2}{4} + \frac{y^2}{3} = 1$ Here, $a^2 = 4$ and $b^2 = 3$. Since $a^2 > b^2$, the major axis is along the x-axis.

2. Find the foci of the ellipse. For an ellipse, the distance of the foci from the center, $c$, is given by the relation $c^2 = a^2 - b^2$. Let the eccentricity of the ellipse be $e_e$. The foci are located at $(\pm ae_e, 0)$ or $(\pm c, 0)$. $c^2 = 4 - 3 = 1$ $c = \sqrt{1} = 1$ So, the foci of the ellipse are at $(\pm 1, 0)$.

3. Use the confocal property. The problem states that the hyperbola is confocal with the ellipse. This means they share the same foci. Therefore, the foci of the hyperbola are also at $(\pm 1, 0)$. Let the equation of the hyperbola be $\frac{x^2}{A^2} - \frac{y^2}{B^2} = 1$ (since the foci are on the x-axis). The distance of the foci of the hyperbola from the center is given by $C$, where $C^2 = A^2 + B^2$. From the foci, we have $C=1$. So, $A^2 + B^2 = 1^2 = 1$

4. Use the information about the hyperbola's transverse axis. The length of the transverse axis of the hyperbola is given as $2\sin \theta$. The length of the transverse axis for our standard hyperbola is $2A$. Therefore, we have: $2A = 2\sin \theta$ $A = \sin \theta$ $A^2 = \sin^2 \theta$

5. Determine the equation of the hyperbola. Now we can find $B^2$ using the relation from step 3: $A^2 + B^2 = 1$ $\sin^2 \theta + B^2 = 1$ $B^2 = 1 - \sin^2 \theta$ Using the trigonometric identity $\sin^2 \theta + \cos^2 \theta = 1$, we get: $B^2 = \cos^2 \theta$ Now, substitute the values of $A^2$ and $B^2$ into the standard equation of the hyperbola: $\frac{x^2}{A^2} - \frac{y^2}{B^2} = 1$ $\frac{x^2}{\sin^2 \theta} - \frac{y^2}{\cos^2 \theta} = 1$

6. Rewrite the equation to match the options. Using the reciprocal trigonometric identities, $\csc\theta = \frac{1}{\sin\theta}$ and $\sec\theta = \frac{1}{\cos\theta}$, we can rewrite the equation as: $x^2 (\frac{1}{\sin^2 \theta}) - y^2 (\frac{1}{\cos^2 \theta}) = 1$ $x^2 \csc^2 \theta - y^2 \sec^2 \theta = 1$ This is often written as $x^2 \text{cosec}^2 \theta - y^2 \sec^2 \theta = 1$.

Comparing this result with the given options, we find that it matches option A.

Conclusion:

The equation of the hyperbola is ${x^2}\cos e{c^2}\theta - {y^2}{\sec ^2}\theta = 1$.