$2sin^{2}B + 4cos\left (A +B \right )sinAsinB + cos2(A+B)$ =

$2sin^{2}B + 4cos\left (A +B \right )sinAsinB + cos2(A+B)$
$\Rightarrow 2\sin^2B + 4\cos (A+B)\sin A. \sin B+2\cos^2(A+B) -1$ [by using the relation of $\Right\cos 2\theta =\cos^2\theta -1$]
$\Rightarrow 2\sin^2B-1 +2\cos (A+B)[2\sin A.\sin B+\cos (A+B)]$
[using the formula of $\cos (A+B) = \cos A.\cos B-\sin A.\sin B$]

$\Rightarrow 2\sin^2B-1 +2\cos (A+B)[2\sin A.\sin B+\cos A.\cos B -\sin A.\sin B]$
$\Rightarrow 2\sin^2B-1 +2\cos (A+B)(\cos (A-B))$
$\Rightarrow 2\sin^2B-1 +2\cos^2A -2\sin^2B$
$\Rightarrow 2\cos^2A-1 = \cos 2A$

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