Question Number 192582 by mathdave last updated on 21/May/23
Commented by mathdave last updated on 21/May/23
$${aAnyone}\:{to}\:{help}\:{me}\:{with}\:{this} \\ $$
Commented by Frix last updated on 21/May/23
$$\mathrm{It}\:\mathrm{can}'\mathrm{t}\:\mathrm{be}\:\mathrm{proven}\:\mathrm{because}\:\mathrm{it}'\mathrm{s}\:\mathrm{wrong}. \\ $$$$\mathrm{See}\:\mathrm{question}\:\mathrm{192466} \\ $$
Commented by York12 last updated on 22/May/23
$${i}\:{guess}\:{you}\:{mean}\:{find}\:\Phi\:{cause}\:{that}\:{is}\:{not} \\ $$$${an}\:{identity} \\ $$$$ \\ $$
Commented by Frix last updated on 22/May/23
$$\mathrm{I}\:\mathrm{showed}\:\mathrm{there}\:\mathrm{is}\:\mathrm{no}\:\mathrm{solution}. \\ $$
Commented by York12 last updated on 22/May/23
$${yeah}\:{you}\:{are}\:{right} \\ $$
Answered by Rajpurohith last updated on 26/May/23
$$\left({sin}^{\mathrm{4}} \theta−{cos}^{\mathrm{4}} \theta\right).{cosec}^{\mathrm{2}} \theta=\frac{\left({sin}^{\mathrm{2}} \theta+{cos}^{\mathrm{2}} \theta\right)\left({sin}^{\mathrm{2}} \theta−{cos}^{\mathrm{2}} \theta\right)}{{sin}^{\mathrm{2}} \theta} \\ $$$$=−\frac{{cos}\left(\mathrm{2}\theta\right)}{{sin}^{\mathrm{2}} \theta}=\frac{−\left(\mathrm{1}−\mathrm{2}{sin}^{\mathrm{2}} \theta\right)}{{sin}^{\mathrm{2}} \theta}=\mathrm{2}−\frac{\mathrm{1}}{{sin}^{\mathrm{2}} \theta}\neq\mathrm{2}\:{for}\:{any}\:\theta. \\ $$