Solution :
L.H.S. = \(2sin2xcos3x + sin2x\over{2cos3x.cos2x + 2cos3x + 2cos^2x}\)
= \(sin2x[2cos3x+1]\over {2[cos3x(cos2x+1)+(cos^2x)]}\)
= \(sin2x[2cos3x+1]\over {2[cos3x(2cos^2x)+(cos^2x)]}\)
= \(sin2x[2cos3x+1]\over {2cos^2x(2cos3x+1)}\) = tanx
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