If $$\dfrac{\cos^{2}48^{o}-\sin^{2}12^{o}}{\sin^{2}24^{o}-\sin^{2}6^{o}}=\dfrac{\alpha+\beta\sqrt{5}}{2}$$, where $$\alpha, \beta \text{ }\epsilon \text{ }N$$, then $$\alpha + \beta $$ is equal to ________

We know that $$\cos\theta=\sin(90-\theta)$$

Thus, $$\cos 48^\circ=\sin (90-48)^\circ=\sin 42^\circ$$

$$\dfrac{\sin^{2}42^{o}-\sin^{2}12^{o}}{\sin^{2}24^{o}-\sin^{2}6^{o}}$$

$$\dfrac{\left(\sin42^o-\sin12^o\right)\left(\sin42^o+\sin12^o\right)}{\left(\sin24^o-\sin6^o\right)\left(\sin24^o+\sin6^o\right)}$$

Now, we know that - 

$$\sin A-\sin B=2\cos\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)$$

$$\sin A+\sin B=2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)$$

Thus, the given equation will become - 

$$\dfrac{2\cos27^o\sin15^o\times2\sin27^o\cos15^o}{2\cos15^o\sin9^o\times2\sin15^o\cos9^o}$$  (Also, $$2\sin A\cos A = \sin 2A$$) 

$$\dfrac{\sin54^o\times\sin30^o}{\sin30^o\times\sin18^o}$$

$$\dfrac{\sin54^o}{\sin18^o}$$

Also, $$\sin54^o=\dfrac{\sqrt{5}+1}{4}$$ and $$\sin18^o=\dfrac{\sqrt{5}-1}{4}$$. Therefore, the fraction becomes - 

$$\dfrac{\frac{\sqrt{5}+1}{4}}{\frac{\sqrt{5}-1}{4}}=\dfrac{\sqrt{5}+1}{\sqrt{5}-1}\times\dfrac{\sqrt{5}+1}{\sqrt{5}+1}=\dfrac{\left(\sqrt{5}+1\right)^2}{4}=\dfrac{6+2\sqrt{5}}{4}=\dfrac{3+\sqrt{5}}{2}$$

Thus, $$\alpha=3$$ and $$\beta=1$$

$$\alpha+\beta=3+1=4$$