Solution :
L.H.S. = 2sin2xcos3x+sin2x2cos3x.cos2x+2cos3x+2cos2x
= sin2x[2cos3x+1]2[cos3x(cos2x+1)+(cos2x)]
= sin2x[2cos3x+1]2[cos3x(2cos2x)+(cos2x)]
= sin2x[2cos3x+1]2cos2x(2cos3x+1) = tanx
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