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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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