Mark the Correct alternative in the following:
The value of 2 sin2 B + 4 cos (A + B) sin A sin B + cos 2 (A + B) is
Given expression is
2 sin2 B + 4 cos (A + B) sin A sin B + cos 2 (A + B)
[ using the cos (A+B) = cos A cos B – sin A sin B]
= 2 sin2 B + 4 sin A sin B [cos A cos B - sin A sin B] + cos 2 (A + B)
= 2 sin2 B + 4 sin A sin B cos A cos B - 4 sin A sin B sin A sin B + cos 2 (A + B)
= 2 sin2 B + (2 sin A cos A) (2sin B cos B) - 4 sin2A sin2 B + cos 2 (A + B)
[ using sin 2A = 2 sin A cos A]
= 2 sin2 B + sin 2A sin 2B - 4 sin2A sin2 B + cos (2A + 2B)
=2 sin2 B ( 1 - 2 sin2A ) + sin 2A sin 2B + (cos2A cos2B - sin 2A sin 2B)
[ using cos (A+B) = cos A cos B – sin A sin B]
=2 sin2 B ( 1 - 2 sin2A )+ sin 2A sin 2B + cos2A cos2B - sin 2A sin 2B
[ using cos 2A = 1 – 2 sin2 x ]
= 2 sin2 B cos 2A + cos2A cos2B
= cos 2A ( 2sin2 B + cos 2B)
[ using cos 2A = cos2 x – sin2 x ]
= cos 2A ( 2 sin2 B + cos2 B – sin2 B)
= cos 2A ( sin2 B + cos2 B)
[using the identity sin2 x + cos2 x = 1]
= cos 2A (1)
= cos 2A
Hence
2 sin2 B + 4 cos (A + B) sin A sin B + cos 2 (A + B) = cos 2A
The answer is option C.