Question Number 47836 by Aknabob1 last updated on 15/Nov/18
$${prove}\:{cosec}^{\mathrm{2}} \theta−{cot}\theta{cosec}\theta=\mathrm{1} \\ $$
Answered by $@ty@m last updated on 15/Nov/18
$${LHS}={cosec}^{\mathrm{2}} \theta−{cot}\theta{cosec}\theta \\ $$$$={cosec}^{\mathrm{2}} \theta−\frac{\mathrm{cos}\:\theta}{\mathrm{sin}\:\theta}×\frac{\mathrm{1}}{\mathrm{sin}\:\theta} \\ $$$$=\frac{\mathrm{1}}{\mathrm{sin}\:^{\mathrm{2}} \theta}−\frac{\mathrm{cos}\:\theta}{\mathrm{sin}\:^{\mathrm{2}} \theta} \\ $$$$=\frac{\mathrm{1}−\mathrm{cos}\:\theta}{\mathrm{sin}\:^{\mathrm{2}} \theta} \\ $$$$=\frac{\mathrm{1}−\mathrm{cos}\:\theta}{\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \theta} \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}+\mathrm{cos}\:\theta} \\ $$$$\neq\mathrm{1} \\ $$$${The}\:{question}\:{is}\:{wrong}. \\ $$$${May}\:{be}\:{typo}\:{error}. \\ $$$${Pl}.\:{check} \\ $$
Commented by Aknabob1 last updated on 15/Nov/18
$${thanks}\:{i}\:{appreciate} \\ $$
Answered by peter frank last updated on 15/Nov/18
$$\mathrm{cosec}\theta\left(\mathrm{cosec}\theta−\mathrm{cot}\theta\right) \\ $$$$\mathrm{cosec}\theta\left(\frac{\mathrm{1}}{\mathrm{sin}\theta}−\frac{\mathrm{cos}\theta}{\mathrm{sin}\theta}\right) \\ $$$$\frac{\mathrm{cosec}\theta}{\mathrm{sin}\theta}\left(\mathrm{1}−\mathrm{cos}\theta\right) \\ $$$$\frac{\mathrm{1}−\mathrm{cos}\theta}{\mathrm{sin}^{\mathrm{2}} \theta}=\frac{\mathrm{1}−\mathrm{cos}\theta}{\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \theta} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}−\mathrm{cos}\theta}{\mathrm{1}+\mathrm{cos}\theta} \\ $$$$\mathrm{hence}\:\:{cosec}^{\mathrm{2}} \theta−{cot}\theta{cosec}\theta\neq\mathrm{1} \\ $$$$ \\ $$
Commented by Aknabob1 last updated on 15/Nov/18
$${thanks}\:{i}\:{appreciate} \\ $$