What is the value of the expression cos2Acos2B + sin²(A - B) - sin²(A + B)?

A. sin(2A - 2B)

B. sin(2A + 2B)

C. cos(2A + 2B)

D. cos(2A - 2B)

Answer: Option C

Solution(By Examveda Team)

cos2Acos2B + sin²(A - B) - sin²(A + B)
= cos2Acos2B + [{sin(A - B) + sin(A + B)}{sin(A - B) - sin(A + B)}]
= cos2Acos2B + [(sinAcosB - cosAsinB + sinAcosB + cosAsinB)(sinAcosB - cosAsinB - sinAcosB - cosAsinB)]
= cos2Acos2B + [(2sinAcosB) × (-2cosAsinB)]
= cos2Acos2B - (2sinAcosA) × (2sinBcosB)
= cos2Acos2B - sin2Asin2B
= cos(2A + 2B)

Alternate:
cos2Acos2B + sin²(A - B) - sin²(A + B)
= cos2Acos2B + sin2Asin2B
= cos(2A + 2B)