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$\cos (36^{0} - A) \cos (36^{0} + A) + \cos (54^{0} + A) \cos (54^{0} - A) =$

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\sin 2 A

\cos 2 A

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detailed solution

Correct option is C

$\cos (36^{0} - A) \cos (36^{0} + A) + \cos (54^{0} + A) \cos (54^{0} - A)$

$= \cos^{2} 36 ° - \sin^{2} A + \cos^{2} 54 ° - \sin^{2} A$

$= \cos^{2} 36 ° + \cos^{2} 54 ° - 2 \sin^{2} A$

$= \cos^{2} 36 ° + \sin^{2} 36 ° - 2 \sin^{2} A$

$= 1 - 2 \sin^{2} A$

$= \cos 2 A$

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$\cos (36^{0} - A) \cos (36^{0} + A) + \cos (54^{0} + A) \cos (54^{0} - A) =$