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We have $\frac{\cos^{2}48^{\circ}-\sin^{2}12^{\circ}}{\sin^{2}24^{\circ}-\sin^{2}6^{\circ}}$. Applying the identity $\cos^2 A - \sin^2 B = \cos(A+B)\cos(A-B)$ to the numerator, we get: $\cos^{2}48^{\circ}-\sin^{2}12^{\circ} = \cos(48^{\circ}+12^{\circ})\cos(48^{\circ}-12^{\circ}) = \cos(60^{\circ})\cos(36^{\circ})$
Applying the identity $\sin^2 A - \sin^2 B = \sin(A+B)\sin(A-B)$ to the denominator, we get: $\sin^{2}24^{\circ}-\sin^{2}6^{\circ} = \sin(24^{\circ}+6^{\circ})\sin(24^{\circ}-6^{\circ}) = \sin(30^{\circ})\sin(18^{\circ})$
Now, we have: $\frac{\cos^{2}48^{\circ}-\sin^{2}12^{\circ}}{\sin^{2}24^{\circ}-\sin^{2}6^{\circ}} = \frac{\cos(60^{\circ})\cos(36^{\circ})}{\sin(30^{\circ})\sin(18^{\circ})}$ We know that $\cos(60^{\circ}) = \frac{1}{2}$ and $\sin(30^{\circ}) = \frac{1}{2}$. So, the expression becomes: $\frac{\frac{1}{2}\cos(36^{\circ})}{\frac{1}{2}\sin(18^{\circ})} = \frac{\cos(36^{\circ})}{\sin(18^{\circ})}$
We know that $\cos(36^{\circ}) = \frac{\sqrt{5}+1}{4}$ and $\sin(18^{\circ}) = \frac{\sqrt{5}-1}{4}$. Therefore, $\frac{\cos(36^{\circ})}{\sin(18^{\circ})} = \frac{\frac{\sqrt{5}+1}{4}}{\frac{\sqrt{5}-1}{4}} = \frac{\sqrt{5}+1}{\sqrt{5}-1}$
To rationalize the denominator, we multiply the numerator and denominator by the conjugate of the denominator, which is $\sqrt{5}+1$: $\frac{\sqrt{5}+1}{\sqrt{5}-1} \times \frac{\sqrt{5}+1}{\sqrt{5}+1} = \frac{(\sqrt{5}+1)^{2}}{(\sqrt{5})^{2}-1^{2}} = \frac{5+2\sqrt{5}+1}{5-1} = \frac{6+2\sqrt{5}}{4} = \frac{3+\sqrt{5}}{2}$
We are given that $\frac{\cos^{2}48^{\circ}-\sin^{2}12^{\circ}}{\sin^{2}24^{\circ}-\sin^{2}6^{\circ}}=\frac{\alpha+\sqrt{5}\beta}{2}$. Comparing $\frac{3+\sqrt{5}}{2}$ with $\frac{\alpha+\sqrt{5}\beta}{2}$, we get $\alpha = 3$ and $\beta = 1$.
We need to find $(\alpha+\beta)$. $\alpha+\beta = 3+1 = 4$
Final Answer: 4