The value of cos (36° − A) cos (36° + A) + cos (54° + A) cos (54° − A) is

(a) sin 2A
(b) cos 2A
(c) cos 3A
(d) sin 3A

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Solution

(b) cos 2A

$\cos (36 ° - A) \cos (36 ° + A) + \cos (54 ° + A) \cos (54 ° - A) = \cos (36 ° - A) \cos (36 ° + A) + \sin [90 ° - (54 ° + A)] \sin [90 ° - (54 ° - A)] [Since \sin (90 ° - θ) = \cos θ] = \cos (36 ° - A) \cos (36 ° + A) + \sin (36 ° - A) \sin (36 ° + A) = \cos (36 ° + A - 36 ° + A) [Using \cos (A - B) formula] = \cos 2 A$

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