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Question Number 10743 by okhema last updated on 24/Feb/17

show that sec^2 θ=((cosec θ)/(cosec θ−sin ))

$${show}\:{that}\:\mathrm{sec}\:^{\mathrm{2}} \theta=\frac{\mathrm{cosec}\:\theta}{\mathrm{cosec}\:\theta−\mathrm{sin}\:} \\ $$

Commented by ridwan balatif last updated on 24/Feb/17

((cosecθ)/(cosecθ−sinθ))=((1/(sinθ))/((1/(sinθ))−sinθ)) =((1/(sinθ))/((1−sin^2 θ)/(sinθ))) =(((1/(sinθ))/(cos^2 θ))/(sinθ)) =(1/(sinθ))×((sinθ)/(cos^2 θ)) =(1/(cos^2 θ)) ((cosecθ)/(cosecθ−sinθ))=sec^2 θ remember! secθ=(1/(cosθ)) cosecθ=(1/(sinθ))

$$\frac{\mathrm{cosec}\theta}{\mathrm{cosec}\theta−\mathrm{sin}\theta}=\frac{\frac{\mathrm{1}}{\mathrm{sin}\theta}}{\frac{\mathrm{1}}{\mathrm{sin}\theta}−\mathrm{sin}\theta}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\frac{\mathrm{1}}{\mathrm{sin}\theta}}{\frac{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \theta}{\mathrm{sin}\theta}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\frac{\frac{\mathrm{1}}{\mathrm{sin}\theta}}{\mathrm{cos}^{\mathrm{2}} \theta}}{\mathrm{sin}\theta} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{sin}\theta}×\frac{\mathrm{sin}\theta}{\mathrm{cos}^{\mathrm{2}} \theta} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \theta} \\ $$$$\frac{\mathrm{cosec}\theta}{\mathrm{cosec}\theta−\mathrm{sin}\theta}=\mathrm{sec}^{\mathrm{2}} \theta \\ $$$$\mathrm{remember}! \\ $$$$\mathrm{sec}\theta=\frac{\mathrm{1}}{\mathrm{cos}\theta} \\ $$$$\mathrm{cosec}\theta=\frac{\mathrm{1}}{\mathrm{sin}\theta} \\ $$

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^2 θ=1+tan^2 θ but thanks anyways

$${yes}\:{thats}\:{what}\:{i}\:{got}\:{but}\:{wasnt}\:{sure}\:{if}\:{its}\:{correct} \\ $$$${cause}\:{i}\:{only}\:{know}\:{sec}^{\mathrm{2}} \theta=\mathrm{1}+{tan}^{\mathrm{2}} \theta\:{but}\:{thanks}\:{anyways} \\ $$

Answered by ridwan balatif last updated on 24/Feb/17

sec^2 θ=(1/(cos^2 θ)) =((1/(1−sin^2 θ)))(((cosecθ)/(cosecθ))) =((cosecθ)/(cosecθ−cosecθ×sinθ×sinθ)) =((cosecθ)/(cosecθ−(1/(sinθ))×sinθ×sinθ)) sec^2 θ=((cosecθ)/(cosecθ−sinθ)) PROVED !!!!

$$\mathrm{sec}^{\mathrm{2}} \theta=\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \theta} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\left(\frac{\mathrm{1}}{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \theta}\right)\left(\frac{\mathrm{cosec}\theta}{\mathrm{cosec}\theta}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{cosec}\theta}{\mathrm{cosec}\theta−\mathrm{cosec}\theta×\mathrm{sin}\theta×\mathrm{sin}\theta} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{cosec}\theta}{\mathrm{cosec}\theta−\frac{\mathrm{1}}{\mathrm{sin}\theta}×\mathrm{sin}\theta×\mathrm{sin}\theta} \\ $$$$\mathrm{sec}^{\mathrm{2}} \theta=\frac{\mathrm{cosec}\theta}{\mathrm{cosec}\theta−\mathrm{sin}\theta} \\ $$$$\mathrm{PROVED}\:!!!! \\ $$$$ \\ $$

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 (1/(cos^2 θ ))

$${why}\:{couldnt}\:{you}\:{start}\:{from}\:{the}\:{right}...{cause}\:{thats}\:{what}\:{i}\:{do}\:{by}\:{doing}\:{the}\:{right}\:{hand}\:{side}\:{but}\:{i}\:{end}\:{up}\:{with}\:\frac{\mathrm{1}}{{cos}^{\mathrm{2}} \theta\:} \\ $$

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