Question Number 192466 by Engr_Jidda last updated on 18/May/23
Commented by Frix last updated on 19/May/23
$$\left(\mathrm{sin}^{\mathrm{4}} \:\theta\:−\mathrm{cos}^{\mathrm{4}} \:\theta\right)\mathrm{csc}^{\mathrm{2}} \:\theta\:=\mathrm{2} \\ $$$$\frac{\mathrm{2sin}^{\mathrm{2}} \:\theta\:−\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \:\theta}=\mathrm{2} \\ $$$$\mathrm{2sin}^{\mathrm{2}} \:\theta\:−\mathrm{1}=\mathrm{2sin}^{\mathrm{2}} \:\theta \\ $$$$−\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{Wrong}.\:\nexists\theta:\:\left(\mathrm{sin}^{\mathrm{4}} \:\theta\:−\mathrm{cos}^{\mathrm{4}} \:\theta\right)\mathrm{csc}^{\mathrm{2}} \:\theta\:=\mathrm{2} \\ $$
Answered by cortano12 last updated on 19/May/23
$$\:\Rightarrow\left(\mathrm{sin}\:^{\mathrm{2}} \theta−\mathrm{cos}\:^{\mathrm{2}} \theta\right).\mathrm{1}.\frac{\mathrm{1}}{\mathrm{sin}\:^{\mathrm{2}} \theta} \\ $$$$\Rightarrow\left(\mathrm{sin}\:^{\mathrm{2}} \theta−\left(\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \theta\right)\right).\frac{\mathrm{1}}{\mathrm{sin}\:^{\mathrm{2}} \theta} \\ $$$$\Rightarrow\left(\mathrm{2sin}\:^{\mathrm{2}} \theta−\mathrm{1}\right).\frac{\mathrm{1}}{\mathrm{sin}\:^{\mathrm{2}} \theta} \\ $$$$\Rightarrow\mathrm{2}−\mathrm{csc}\:^{\mathrm{2}} \theta\: \\ $$
Answered by Spillover last updated on 19/May/23
![(sin^2 θ+cos^2 θ)(sin^2 θ−cos^2 θ)]cosec^2 θ (sin^2 θ−cos^2 θ)]cosec^2 θ 1−cot^2 θ please check](https://www.tinkutara.com/question/Q192473.png)
$$\left.\left(\mathrm{sin}\:^{\mathrm{2}} \theta+\mathrm{cos}\:^{\mathrm{2}} \theta\right)\left(\mathrm{sin}\:^{\mathrm{2}} \theta−\mathrm{cos}\:^{\mathrm{2}} \theta\right)\right]\mathrm{cosec}\:^{\mathrm{2}} \theta \\ $$$$\left.\left(\mathrm{sin}\:^{\mathrm{2}} \theta−\mathrm{cos}\:^{\mathrm{2}} \theta\right)\right]\mathrm{cosec}\:^{\mathrm{2}} \theta \\ $$$$\mathrm{1}−\mathrm{cot}\:^{\mathrm{2}} \theta \\ $$$${please}\:{check} \\ $$