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What is the value of the expression​cos⁡2Acos⁡2B+sin⁡2(A−B)−sin⁡2(A+B)\cos 2A \cos 2B + \sin^2 (A - B) - \sin^2 (A + B)cos2Acos2B+sin2(A−B)−sin2(A+B)​
Question

What is the value of the expression

cos2Acos2B+sin2(AB)sin2(A+B)\cos 2A \cos 2B + \sin^2 (A - B) - \sin^2 (A + B)​​

A.

sin (2A-2B)

B.

sin (2A+2B)

C.

cos(2A+2B)

D.

cos(2A-2B)

Correct option is C

Given: 
To find: cos2Acos2B+sin2(AB)sin2(A+B)\cos 2A\cos2B+\sin^2(A-B)-sin^2(A+B) 
Formula Used:  
cos(2A+2B)=cos2Acos2Bsin2Asin2Bsin2(AB)sin2(A+B)=sin(2A)sin(2B)\cos (2A + 2B) = \cos2A \cos 2B -\sin2A \sin2B \\\sin^2(A - B) - \sin^2(A + B) = -\sin(2A)\sin(2B) 
Solution:
=cos2Acos2B+sin(AB+A+B)sin(ABAB)=cos2Acos2B+sin2A×sin2B=cos2Acos2Bsin2Asin2B=cos2Acos2Bsin2Asin2B=cos(2A+2B)= \cos 2A \cos 2B + \sin (A - B + A + B) \sin (A - B - A - B) \\= \cos 2A \cos 2B + \sin 2A \times -\sin 2B \\= \cos 2A \cos 2B - \sin 2A \sin 2B \\= \cos 2A \cos 2B - \sin 2A \sin 2B \\= \cos (2A + 2B) 
Thus, correct option is (c).

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