Question : What is the value of the expression $\cos 2 A \cos 2 B+\sin ^2(A-B)-\sin ^2(A+B)$?
Option 1: $\sin (2 A-2 B)$
Option 2: $\sin (2 A+2 B)$
Option 3: $\cos (2 A+2 B)$
Option 4: $\cos (2 A-2 B)$
Correct Answer: $\cos (2 A+2 B)$
Solution :
Given, $\cos(2A)\cos(2B)+\sin^2(A-B)-\sin^2(A+B)$
$=\cos(2A)\cos(2B) - [\sin^2(A + B) - \sin^2(A - B)]$
$=\cos(2A)\cos(2B) - [\sin(A + B) + \sin(A - B)][\sin(A + B) - \sin(A - B)]$
Using Identities: $\sin C + \sin D = 2\sin \frac{C+D}{2}\cos \frac{C-D}{2}$ and
$\sin C - \sin D = 2\cos \frac{C+D}{2}\sin \frac{C-D}{2}$, we get,
$=\cos(2A)\cos(2B) - [2\sin \frac{(A + B)+(A-B)}{2}\cos \frac{(A + B)-(A-B)}{2}][2\cos \frac{(A + B)+(A-B)}{2}\sin \frac{(A + B)-(A-B)}{2}]$
$=\cos(2A)\cos(2B) - (2\sin A\cos B)(2\cos A\sin B)$
$=\cos{2A}\cos{2B}-\sin{2A}\sin{2B}$
$=\cos (2A + 2B)$
Hence, the correct answer is $\cos (2A + 2B)$.
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