$(L_1L_2)^R = \{ x^R | x \in L_1L_2 \}$
Since $x \in L_1L_2,$ hence, $x = yz; \,\, y \in L_1, z \in L_2$
$(L_1L_2)^R = \{ (yz)^R | y \in L_1, z \in L_2 \}$
$(L_1L_2)^R = \{ z^Ry^R | y \in L_1, z \in L_2 \}$ (by the result we proved Here )
$(L_1L_2)^R = \{ z^R | z \in L_2 \}.$ $\{ y^R | y \in L_1 \}$
$(L_1L_2)^R = L_2^R.L_1^R.$
Hence Proved.
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