Question Number 10743 by okhema last updated on 24/Feb/17
$${show}\:{that}\:\mathrm{sec}\:^{\mathrm{2}} \theta=\frac{\mathrm{cosec}\:\theta}{\mathrm{cosec}\:\theta−\mathrm{sin}\:} \\ $$
Commented by ridwan balatif last updated on 24/Feb/17
$$\frac{\mathrm{cosec}\theta}{\mathrm{cosec}\theta−\mathrm{sin}\theta}=\frac{\frac{\mathrm{1}}{\mathrm{sin}\theta}}{\frac{\mathrm{1}}{\mathrm{sin}\theta}−\mathrm{sin}\theta}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\frac{\mathrm{1}}{\mathrm{sin}\theta}}{\frac{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \theta}{\mathrm{sin}\theta}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\frac{\frac{\mathrm{1}}{\mathrm{sin}\theta}}{\mathrm{cos}^{\mathrm{2}} \theta}}{\mathrm{sin}\theta} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{sin}\theta}×\frac{\mathrm{sin}\theta}{\mathrm{cos}^{\mathrm{2}} \theta} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \theta} \\ $$$$\frac{\mathrm{cosec}\theta}{\mathrm{cosec}\theta−\mathrm{sin}\theta}=\mathrm{sec}^{\mathrm{2}} \theta \\ $$$$\mathrm{remember}! \\ $$$$\mathrm{sec}\theta=\frac{\mathrm{1}}{\mathrm{cos}\theta} \\ $$$$\mathrm{cosec}\theta=\frac{\mathrm{1}}{\mathrm{sin}\theta} \\ $$
Commented by okhema last updated on 24/Feb/17
$${yes}\:{thats}\:{what}\:{i}\:{got}\:{but}\:{wasnt}\:{sure}\:{if}\:{its}\:{correct} \\ $$$${cause}\:{i}\:{only}\:{know}\:{sec}^{\mathrm{2}} \theta=\mathrm{1}+{tan}^{\mathrm{2}} \theta\:{but}\:{thanks}\:{anyways} \\ $$
Answered by ridwan balatif last updated on 24/Feb/17
$$\mathrm{sec}^{\mathrm{2}} \theta=\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \theta} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\left(\frac{\mathrm{1}}{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \theta}\right)\left(\frac{\mathrm{cosec}\theta}{\mathrm{cosec}\theta}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{cosec}\theta}{\mathrm{cosec}\theta−\mathrm{cosec}\theta×\mathrm{sin}\theta×\mathrm{sin}\theta} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{cosec}\theta}{\mathrm{cosec}\theta−\frac{\mathrm{1}}{\mathrm{sin}\theta}×\mathrm{sin}\theta×\mathrm{sin}\theta} \\ $$$$\mathrm{sec}^{\mathrm{2}} \theta=\frac{\mathrm{cosec}\theta}{\mathrm{cosec}\theta−\mathrm{sin}\theta} \\ $$$$\mathrm{PROVED}\:!!!! \\ $$$$ \\ $$
Commented by okhema last updated on 24/Feb/17
$${why}\:{couldnt}\:{you}\:{start}\:{from}\:{the}\:{right}…{cause}\:{thats}\:{what}\:{i}\:{do}\:{by}\:{doing}\:{the}\:{right}\:{hand}\:{side}\:{but}\:{i}\:{end}\:{up}\:{with}\:\frac{\mathrm{1}}{{cos}^{\mathrm{2}} \theta\:} \\ $$