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Question Number 47836 by Aknabob1 last updated on 15/Nov/18

prove cosec^2 θ−cotθcosecθ=1

$${prove}\:{cosec}^{\mathrm{2}} \theta−{cot}\theta{cosec}\theta=\mathrm{1} \\ $$

Answered by $@ty@m last updated on 15/Nov/18

LHS=cosec^2 θ−cotθcosecθ  =cosec^2 θ−((cos θ)/(sin θ))×(1/(sin θ))  =(1/(sin^2 θ))−((cos θ)/(sin^2 θ))  =((1−cos θ)/(sin^2 θ))  =((1−cos θ)/(1−cos^2 θ))  =(1/(1+cos θ))  ≠1  The question is wrong.  May be typo error.  Pl. check

$${LHS}={cosec}^{\mathrm{2}} \theta−{cot}\theta{cosec}\theta \\ $$$$={cosec}^{\mathrm{2}} \theta−\frac{\mathrm{cos}\:\theta}{\mathrm{sin}\:\theta}×\frac{\mathrm{1}}{\mathrm{sin}\:\theta} \\ $$$$=\frac{\mathrm{1}}{\mathrm{sin}\:^{\mathrm{2}} \theta}−\frac{\mathrm{cos}\:\theta}{\mathrm{sin}\:^{\mathrm{2}} \theta} \\ $$$$=\frac{\mathrm{1}−\mathrm{cos}\:\theta}{\mathrm{sin}\:^{\mathrm{2}} \theta} \\ $$$$=\frac{\mathrm{1}−\mathrm{cos}\:\theta}{\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \theta} \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}+\mathrm{cos}\:\theta} \\ $$$$\neq\mathrm{1} \\ $$$${The}\:{question}\:{is}\:{wrong}. \\ $$$${May}\:{be}\:{typo}\:{error}. \\ $$$${Pl}.\:{check} \\ $$

Commented by Aknabob1 last updated on 15/Nov/18

thanks i appreciate

$${thanks}\:{i}\:{appreciate} \\ $$

Answered by peter frank last updated on 15/Nov/18

cosecθ(cosecθ−cotθ)  cosecθ((1/(sinθ))−((cosθ)/(sinθ)))  ((cosecθ)/(sinθ))(1−cosθ)  ((1−cosθ)/(sin^2 θ))=((1−cosθ)/(1−cos^2 θ))                =((1−cosθ)/(1+cosθ))  hence  cosec^2 θ−cotθcosecθ≠1

$$\mathrm{cosec}\theta\left(\mathrm{cosec}\theta−\mathrm{cot}\theta\right) \\ $$$$\mathrm{cosec}\theta\left(\frac{\mathrm{1}}{\mathrm{sin}\theta}−\frac{\mathrm{cos}\theta}{\mathrm{sin}\theta}\right) \\ $$$$\frac{\mathrm{cosec}\theta}{\mathrm{sin}\theta}\left(\mathrm{1}−\mathrm{cos}\theta\right) \\ $$$$\frac{\mathrm{1}−\mathrm{cos}\theta}{\mathrm{sin}^{\mathrm{2}} \theta}=\frac{\mathrm{1}−\mathrm{cos}\theta}{\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \theta} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}−\mathrm{cos}\theta}{\mathrm{1}+\mathrm{cos}\theta} \\ $$$$\mathrm{hence}\:\:{cosec}^{\mathrm{2}} \theta−{cot}\theta{cosec}\theta\neq\mathrm{1} \\ $$$$ \\ $$

Commented by Aknabob1 last updated on 15/Nov/18

thanks i appreciate

$${thanks}\:{i}\:{appreciate} \\ $$

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