The value of $2 \sin^{2} B + 4 \cos (A + B) \sin A \sin B + \cos 2 (A + B)$ is
(a) 0
(b) cos 3A
(c) cos 2A
(d) none of these

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Solution

(c) cos 2A

$We have, 2 \sin^{2} B + 4 \cos (A + B) \sin A \sin B + \cos 2 (A + B) = 1 - \cos 2 B + \cos 2 (A + B) + 4 \cos (A + B) \sin A \sin B = 1 + (\cos 2 (A + B) - \cos 2 B) + 4 \cos (A + B) \sin A \sin B = 1 - 2 \sin A \sin (A + 2 B) + 4 \cos (A + B) \sin A \sin B [∵ \cos C - \cos D = - 2 \sin \frac{C + D}{2} \sin \frac{C - D}{2}] = 1 - 2 \sin A [\sin (A + 2 B) - 2 \sin B \cos (A + B)] = 1 - 2 \sin A [\sin (A + 2 B) - \{\sin (B + A + B) + \sin (B - (A + B))\}] [∵ 2 \sin C \cos D = \sin (C + D) + \sin (C - D)] = 1 - 2 \sin A [\sin (A + 2 B) - \{\sin (A + 2 B) + \sin (- A)\}] = 1 - 2 \sin A [\sin A] = 1 - 2 \sin^{2} A = \cos 2 A$

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