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Q.

sin23Asin2A-cos2Acos2A=

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a

8cos2A

b

4cos2A

c

2cos2A

d

cos2A

answer is D.

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Detailed Solution

  sin23Asin2A-cos23Acos2A =sin23A cos2A-cos23A sin2Asin2A cos2A =(sin3A cos A-cos3A sin A) (sin3A cos A+cos3A sin A)sin2A cos2A =sin2A sin4Asin2A cos2A =2sin A cos A 2sin2A cos2Asin2A cos2A =4sinA cos A 2sin A cos A cos2Asin2A cos2A =8cos2A

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