The value of cos(18^{circ}-A) cos(18^{circ}+A | 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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$\cos 36^{\circ}$

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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 $\cos(90^{\circ}-	heta) = \sin 	heta$: $\cos(72^{\circ}-A) = \sin(18^{\circ}+A)$ and $\cos(72^{\circ}+A) = \sin(18^{\circ}-A)$ Expression $= \cos(18^{\circ}-A)\cos(18^{\circ}+A) - \sin(18^{\circ}+A)\sin(18^{\circ}-A)$ $= \cos((18^{\circ}-A) + (18^{\circ}+A)) = \cos 36^{\circ}$

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

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