The value of cos(18^{circ}-A) cdot cos(18^{ci | vedclass

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(B) Using the identity $\cos(x-y)\cos(x+y) = \cos^2 x - \sin^2 y$,we have: $\cos(18^{\circ}-A)\cos(18^{\circ}+A) = \cos^2 18^{\circ} - \sin^2 A$ $\cos

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$\sin 54^{\circ}$

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(B) Using the identity $\cos(x-y)\cos(x+y) = \cos^2 x - \sin^2 y$,we have: $\cos(18^{\circ}-A)\cos(18^{\circ}+A) = \cos^2 18^{\circ} - \sin^2 A$ $\cos(72^{\circ}-A)\cos(72^{\circ}+A) = \cos^2 72^{\circ} - \sin^2 A$ Subtracting these: $(\cos^2 18^{\circ} - \sin^2 A) - (\cos^2 72^{\circ} - \sin^2 A) = \cos^2 18^{\circ} - \cos^2 72^{\circ}$ Since $\cos 72^{\circ} = \sin 18^{\circ}$,this becomes $\cos^2 18^{\circ} - \sin^2 18^{\circ} = \cos(2 	imes 18^{\circ}) = \cos 36^{\circ}$. Alternatively,using $\cos(x-y)\cos(x+y) = \frac{1}{2}(\cos 2x + \cos 2y)$: $\frac{1}{2}(\cos 36^{\circ} + \cos 2A) - \frac{1}{2}(\cos 144^{\circ} + \cos 2A) = \frac{1}{2}(\cos 36^{\circ} - \cos 144^{\circ})$ Since $\cos 144^{\circ} = -\cos 36^{\circ}$,we get $\frac{1}{2}(\cos 36^{\circ} + \cos 36^{\circ}) = \cos 36^{\circ} = \sin 54^{\circ}$.

The value of $\cos(18^{\circ}-A) \cdot \cos(18^{\circ}+A) - \cos(72^{\circ}-A) \cdot \cos(72^{\circ}+A)$ is:

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