cosec θcosec θ−1+cosec θcosec θ+1−tan2 θ\frac{cosec \ \theta}{cosec \ \theta - 1}+\frac{cosec \ \theta}{cosec \ \theta + 1}-tan^2 \ \thetacosec θ−1cosec θ+cosec θ+1cosec θ−tan2 θ, 0º < Θ < 90º, is equal to: Asec2 Θ + 1 Bsec2 Θ C2sec2 Θ D1 - tan2Θ Show answer & explanationCorrect answer: AExplanationThe correct answer is option 1 i.e. sec2 Θ + 1 cosec θcosec θ−1+cosec θcosec θ+1−tan2 θ\frac{cosec \ \theta}{cosec \ \theta - 1}+\frac{cosec \ \theta}{cosec \ \theta + 1}-tan^2 \ \thetacosec θ−1cosec θ+cosec θ+1cosec θ−tan2 θ cosec θ(cosec θ+1)+cosec θ(cosec θ−1)cosec2 θ − 1−tan2 θ\frac{cosec \ \theta(cosec \ \theta+1)+cosec \ \theta(cosec \ \theta-1)}{cosec^2 \ \theta \ - \ 1}-tan^2 \ \thetacosec2 θ − 1cosec θ(cosec θ+1)+cosec θ(cosec θ−1)−tan2 θ 2cosec2 θcosec2 θ − 1−tan2 θ\frac{2cosec^2 \ \theta}{cosec^2 \ \theta \ - \ 1}-tan^2 \ \thetacosec2 θ − 12cosec2 θ−tan2 θ 2cosec2 θcot2 θ−tan2 θ\frac{2cosec^2 \ \theta}{cot^2 \ \theta}-tan^2 \ \thetacot2 θ2cosec2 θ−tan2 θ 2sec2 θ−tan2 θ2sec^2 \ \theta-tan^2 \ \theta2sec2 θ−tan2 θ 2sec2 θ−sec2 θ+12sec^2 \ \theta-sec^2 \ \theta + 12sec2 θ−sec2 θ+1 sec2 θ+1sec^2 \ \theta + 1sec2 θ+1