Question : Evaluate the following:
$\cos \left(36^{\circ}+A\right) \cdot \cos \left(36^{\circ}-A\right)+\cos \left(54^{\circ}+A\right) \cdot \cos \left(54^{\circ}-A\right)$
Option 1: $\sin 2A$
Option 2: $\cos A$
Option 3: $\sin A$
Option 4: $\cos 2A$
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Correct Answer: $\cos 2A$
Solution :
$\cos (36^{\circ}+A). \cos (36^{\circ}-A)+\cos (54^{\circ}+A) .\cos (54^{\circ}-A)$
= $\cos (36^{\circ}+A) . \cos (36^{\circ}-A)+\sin [90^{\circ}-(54^{\circ}+A)] . \sin [90^{\circ}-(54^{\circ}-A )]$
= $\cos (36^{\circ}+A) \cdot \cos (36^{\circ}-A)+\sin (36^{\circ}-A) \cdot \sin (36^{\circ}+A)$
= $\cos (36^{\circ}+A-36^{\circ}+A )$ [$\because \cos(A- B) =\sin A.\sin B +\cos A. \cos B$]
= $\cos 2A$
Hence, the correct answer is $\cos 2A$.
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