Question Number 226111 by Linton last updated on 19/Nov/25
$${Show}\:{that}\:{the}\:{equation} \\ $$$$\frac{\mathrm{1}}{{sin}\theta+{cos}\theta}\:+\:\frac{\mathrm{1}}{{sin}\theta−{cos}\theta}\:=\:\mathrm{1} \\ $$$${may}\:{be}\:{express}\:{in}\:{the}\:{form} \\ $$$${a}\left({sin}\theta\right)^{\mathrm{2}} +{bsin}\theta+{c}=\mathrm{0}\:{where}\:{a}\:{b}\: \\ $$$${and}\:{c}\:{are}\:{constants}\:{to}\:{be}\:{found}. \\ $$
Answered by Frix last updated on 20/Nov/25
$$\frac{\mathrm{1}}{{s}+{c}}+\frac{\mathrm{1}}{{s}−{c}}=\mathrm{1} \\ $$$$\left({s}−{c}\right)+\left({s}+{c}\right)=\left({s}+{c}\right)\left({s}−{c}\right) \\ $$$$\mathrm{2}{s}={s}^{\mathrm{2}} −{c}^{\mathrm{2}} \\ $$$$\mathrm{2}{s}={s}^{\mathrm{2}} −\left(\mathrm{1}−{s}^{\mathrm{2}} \right) \\ $$$$\mathrm{2}{s}=\mathrm{2}{s}^{\mathrm{2}} −\mathrm{1} \\ $$$$\mathrm{2sin}^{\mathrm{2}} \:\theta\:−\mathrm{2sin}\:\theta\:−\mathrm{1}=\mathrm{0} \\ $$