Question Number 106609 by DeepakMahato last updated on 06/Aug/20
$${Prove}\:{that} \\ $$$$\left(\mathrm{1}−{sin}^{\mathrm{2}} \theta\right){sec}^{\mathrm{2}} \theta=\mathrm{1} \\ $$
Commented by mohammad17 last updated on 06/Aug/20
$${when}:\mathrm{1}−{sin}^{\mathrm{2}} \theta={cos}^{\mathrm{2}} \theta \\ $$$${when}:{sec}^{\mathrm{2}} \theta=\frac{\mathrm{1}}{{cos}^{\mathrm{2}} \theta} \\ $$$$ \\ $$$$\rfloor\left(\mathrm{1}−{sin}^{\mathrm{2}} \theta\right){sec}^{\mathrm{2}} \theta={cos}^{\mathrm{2}} \theta×\frac{\mathrm{1}}{{cos}^{\mathrm{2}} \theta}=\mathrm{1} \\ $$$$ \\ $$$$\:\:\:{m}.{ss}.{mohammad} \\ $$
Answered by JDamian last updated on 06/Aug/20
$$\left(\mathrm{1}−\left(\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \theta\right)\right)\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \theta}=\mathrm{cos}^{\mathrm{2}} \theta\:\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \theta}\:=\:\mathrm{1} \\ $$