Question Number 93571 by prince 5 last updated on 13/May/20
$${Prove}\:{that}\:\frac{\mathrm{1}−{tan}^{\mathrm{3}} \theta}{\mathrm{1}+{tan}^{\mathrm{3}} \theta}\:=\mathrm{1}−\mathrm{2}{sin}^{\mathrm{2}} \theta \\ $$
Commented by mr W last updated on 13/May/20
$${you}\:{can}'{t}\:{prove}\:{something}\:{which}\:{is}\: \\ $$$${false}.\:{just}\:{test}\:{with}\:\theta=\mathrm{30}°. \\ $$