Question Number 85484 by oustmuchiya@gmail.com last updated on 22/Mar/20
$${show}\:{that}\:\frac{\mathrm{1}}{{sec}\theta+\mathrm{1}}+\frac{\mathrm{1}}{{sec}\theta−\mathrm{1}}\equiv\mathrm{2}{cosec}\theta{cot}\theta \\ $$
Answered by som(math1967) last updated on 18/Apr/20
$$\frac{\mathrm{sec}\:\theta−\mathrm{1}+{sec}\theta+\mathrm{1}}{\left({sec}\theta+\mathrm{1}\right)\left({sec}\theta−\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{2}{sec}\theta}{{sec}^{\mathrm{2}} \theta−\mathrm{1}}=\frac{\mathrm{2}{sec}\theta}{{tan}^{\mathrm{2}} \theta}=\frac{\mathrm{2}{sec}\theta{cos}^{\mathrm{2}} \theta}{{sin}^{\mathrm{2}} \theta} \\ $$$$=\frac{\mathrm{2}{cos}\theta}{{sin}\theta.{sin}\theta}=\mathrm{2}{cosec}\theta{cot}\theta \\ $$