Question Number 129073 by amns last updated on 12/Jan/21
$$\boldsymbol{{Prove}}\:\boldsymbol{{that}}: \\ $$$${sec}^{\mathrm{4}} \theta\:−\:{sec}^{\mathrm{2}} \theta\:=\:{tan}^{\mathrm{4}} \theta\:+\:{tan}^{\mathrm{2}} \theta \\ $$$${Please}\:{help}\:{me}\:{quickly}! \\ $$
Commented by benjo_mathlover last updated on 12/Jan/21
$$\mathrm{sec}\:^{\mathrm{4}} \theta−\mathrm{sec}\:^{\mathrm{2}} \theta=\mathrm{sec}\:^{\mathrm{2}} \theta\left(\mathrm{sec}\:^{\mathrm{2}} \theta−\mathrm{1}\right) \\ $$$$\:=\:\left(\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \theta\right)\left(\mathrm{tan}\:^{\mathrm{2}} \theta\right) \\ $$$$\:=\:\mathrm{tan}\:^{\mathrm{2}} \theta+\mathrm{tan}\:^{\mathrm{4}} \theta \\ $$
Answered by MJS_new last updated on 12/Jan/21
$$\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{4}} \:{x}}−\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \:{x}}= \\ $$$$ \\ $$$$=\frac{\mathrm{1}}{{c}^{\mathrm{4}} }−\frac{\mathrm{1}}{{c}^{\mathrm{2}} }=\frac{\mathrm{1}−{c}^{\mathrm{2}} }{{c}^{\mathrm{4}} }=\frac{{s}^{\mathrm{2}} }{{c}^{\mathrm{4}} }=\frac{{s}^{\mathrm{2}} ×\mathrm{1}}{{c}^{\mathrm{4}} }=\frac{{s}^{\mathrm{2}} \left({s}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)}{{c}^{\mathrm{4}} }= \\ $$$$=\frac{{s}^{\mathrm{4}} +{s}^{\mathrm{2}} {c}^{\mathrm{2}} }{{c}^{\mathrm{4}} }=\frac{{s}^{\mathrm{4}} }{{c}^{\mathrm{4}} }+\frac{{s}^{\mathrm{2}} }{{c}^{\mathrm{2}} }= \\ $$$$=\mathrm{tan}^{\mathrm{4}} \:{x}\:+\mathrm{ran}^{\mathrm{2}} \:{x} \\ $$