Question Number 31679 by pieroo last updated on 12/Mar/18
$$\mathrm{prove}\:\mathrm{that}\:\mathrm{sec}^{\mathrm{2}} \theta\:\mathrm{cosec}^{\mathrm{2}} \theta\:=\mathrm{sec}^{\mathrm{2}} \theta+\mathrm{cosec}^{\mathrm{2}} \theta \\ $$
Answered by Tinkutara last updated on 12/Mar/18
$$\mathrm{sec}^{\mathrm{2}} \:\theta\mathrm{cosec}^{\mathrm{2}} \:\theta=\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \:\theta\mathrm{cos}^{\mathrm{2}} \:\theta} \\ $$$$=\frac{\mathrm{sin}^{\mathrm{2}} \:\theta+\mathrm{cos}^{\mathrm{2}} \:\theta\:}{\mathrm{sin}^{\mathrm{2}} \:\theta\mathrm{cos}^{\mathrm{2}} \:\theta} \\ $$$$=\frac{\mathrm{1}\:}{\mathrm{cos}^{\mathrm{2}} \:\theta}+\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \:\theta}=\mathrm{sec}^{\mathrm{2}} \:\theta+\mathrm{cosec}^{\mathrm{2}} \:\theta \\ $$
Commented by pieroo last updated on 12/Mar/18
$$\mathrm{thanks}\:\mathrm{boss} \\ $$
Answered by Joel578 last updated on 12/Mar/18
$$\mathrm{sec}^{\mathrm{2}} \:\theta\:\mathrm{cosec}^{\mathrm{2}} \:\theta\:=\:\left(\mathrm{1}\:+\:\mathrm{tan}^{\mathrm{2}} \:\theta\right)\left(\mathrm{1}\:+\:\mathrm{cot}^{\mathrm{2}} \:\theta\right) \\ $$$$=\:\mathrm{1}\:+\:\mathrm{cot}^{\mathrm{2}} \:\theta\:+\:\mathrm{tan}^{\mathrm{2}} \:\theta\:+\:\mathrm{tan}^{\mathrm{2}} \:\theta\:\mathrm{cot}^{\mathrm{2}} \:\theta \\ $$$$=\:\mathrm{cosec}^{\mathrm{2}} \:\theta\:+\:\mathrm{sec}^{\mathrm{2}} \:\theta \\ $$
Commented by pieroo last updated on 12/Mar/18
$$\mathrm{i}\:\mathrm{am}\:\mathrm{grateful} \\ $$