(tan³ θ - 1) / (tan θ - 1) = sec² θ + tan θ

As, a 3 - b 3 = (a - b)(a 2 + ab + b 2 ) ∴ tan 3 θ - (1) 3 = (tan θ - 1)(tan 2 θ + tan θ + 1) The L.H.S. of the equation can be written as, Since, L.H.S. = R.H.S. hence, proved that = sec 2 θ + tan θ.