L.H.S = tan−1(6x−8x31−12x2)−tan−1(4x1−4x2)
=tan−1[3×2x−(2x)31−3×(2x)2]−tan−1[2×2x1−(2x)2]
Considering tan−12x=θ then 2x=tanθ
Now,
= tan−1[3tan θ−(tan θ)31−3(tan θ)2]−tan−1[2tan θ1−tan2 θ]
= tan−1[3tan θ−tan3 θ1−3 tan2θ]−tan−1[2tan θ1−tan2 θ]
= tan−1(tan 3θ)−tan−1(tan 2θ)
= 3θ−2θ=θ=tan−12x = R.H.S.
Hence proved.