A solution of the equation (1−tan⁡θ)(1+tan⁡θ)sec2⁡θ+2tan2⁡θ=0 which θ lies in the interval (−π/2,π/2) is given by

(1−tan⁡θ)(1+tan⁡θ)sec2⁡θ+2tan2⁡θ=0⇒1−tan4⁡θ+2tan2⁡θ=0⇒2t=t2−1 where t=tan2⁡θBy inspection (or by graph) we findy=2t and y=t2−1 intersect in t=3⇒ tan2⁡θ=3⇒ tan⁡θ=±3⇒θ=±π/3