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

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