If frac{cos ^2 48^{circ}-sin ^2 12^{circ}}{si | vedclass

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(B) Using the identities $\cos ^2 A - \sin ^2 B = \cos(A+B) \cos(A-B)$ and $\sin ^2 A - \sin ^2 B = \sin(A+B) \sin(A-B)$:Numerator: $\cos ^2 48^{\circ

Vedclass · Accepted Answer

$4$

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(B) Using the identities $\cos ^2 A - \sin ^2 B = \cos(A+B) \cos(A-B)$ and $\sin ^2 A - \sin ^2 B = \sin(A+B) \sin(A-B)$:Numerator: $\cos ^2 48^{\circ} - \sin ^2 12^{\circ} = \cos(48^{\circ}+12^{\circ}) \cos(48^{\circ}-12^{\circ}) = \cos 60^{\circ} \cos 36^{\circ}$.Denominator: $\sin ^2 24^{\circ} - \sin ^2 6^{\circ} = \sin(24^{\circ}+6^{\circ}) \sin(24^{\circ}-6^{\circ}) = \sin 30^{\circ} \sin 18^{\circ}$.Substituting the values $\cos 60^{\circ} = \frac{1}{2}$, $\cos 36^{\circ} = \frac{\sqrt{5}+1}{4}$, $\sin 30^{\circ} = \frac{1}{2}$, and $\sin 18^{\circ} = \frac{\sqrt{5}-1}{4}$:Expression $= \frac{(1/2) 	imes ((\sqrt{5}+1)/4)}{(1/2) 	imes ((\sqrt{5}-1)/4)} = \frac{\sqrt{5}+1}{\sqrt{5}-1}$.Rationalizing the denominator: $\frac{(\sqrt{5}+1)(\sqrt{5}+1)}{(\sqrt{5}-1)(\sqrt{5}+1)} = \frac{5+1+2\sqrt{5}}{5-1} = \frac{6+2\sqrt{5}}{4} = \frac{3+\sqrt{5}}{2}$.Comparing with $\frac{\alpha+\beta \sqrt{5}}{2}$, we get $\alpha = 3$ and $\beta = 1$.Therefore, $\alpha+\beta = 3+1 = 4$.

If $\frac{\cos ^2 48^{\circ}-\sin ^2 12^{\circ}}{\sin ^2 24^{\circ}-\sin ^2 6^{\circ}}=\frac{\alpha+\beta \sqrt{5}}{2}$, where $\alpha, \beta \in N$, then $\alpha+\beta$ is equal to . . . . . . .

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