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Question Number 31679 by pieroo last updated on 12/Mar/18

prove that sec^2 θ cosec^2 θ =sec^2 θ+cosec^2 θ

$$\mathrm{prove}\:\mathrm{that}\:\mathrm{sec}^{\mathrm{2}} \theta\:\mathrm{cosec}^{\mathrm{2}} \theta\:=\mathrm{sec}^{\mathrm{2}} \theta+\mathrm{cosec}^{\mathrm{2}} \theta \\ $$

Answered by Tinkutara last updated on 12/Mar/18

sec^2  θcosec^2  θ=(1/(sin^2  θcos^2  θ))  =((sin^2  θ+cos^2  θ )/(sin^2  θcos^2  θ))  =((1 )/(cos^2  θ))+(1/(sin^2  θ))=sec^2  θ+cosec^2  θ

$$\mathrm{sec}^{\mathrm{2}} \:\theta\mathrm{cosec}^{\mathrm{2}} \:\theta=\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \:\theta\mathrm{cos}^{\mathrm{2}} \:\theta} \\ $$$$=\frac{\mathrm{sin}^{\mathrm{2}} \:\theta+\mathrm{cos}^{\mathrm{2}} \:\theta\:}{\mathrm{sin}^{\mathrm{2}} \:\theta\mathrm{cos}^{\mathrm{2}} \:\theta} \\ $$$$=\frac{\mathrm{1}\:}{\mathrm{cos}^{\mathrm{2}} \:\theta}+\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \:\theta}=\mathrm{sec}^{\mathrm{2}} \:\theta+\mathrm{cosec}^{\mathrm{2}} \:\theta \\ $$

Commented by pieroo last updated on 12/Mar/18

thanks boss

$$\mathrm{thanks}\:\mathrm{boss} \\ $$

Answered by Joel578 last updated on 12/Mar/18

sec^2  θ cosec^2  θ = (1 + tan^2  θ)(1 + cot^2  θ)  = 1 + cot^2  θ + tan^2  θ + tan^2  θ cot^2  θ  = cosec^2  θ + sec^2  θ

$$\mathrm{sec}^{\mathrm{2}} \:\theta\:\mathrm{cosec}^{\mathrm{2}} \:\theta\:=\:\left(\mathrm{1}\:+\:\mathrm{tan}^{\mathrm{2}} \:\theta\right)\left(\mathrm{1}\:+\:\mathrm{cot}^{\mathrm{2}} \:\theta\right) \\ $$$$=\:\mathrm{1}\:+\:\mathrm{cot}^{\mathrm{2}} \:\theta\:+\:\mathrm{tan}^{\mathrm{2}} \:\theta\:+\:\mathrm{tan}^{\mathrm{2}} \:\theta\:\mathrm{cot}^{\mathrm{2}} \:\theta \\ $$$$=\:\mathrm{cosec}^{\mathrm{2}} \:\theta\:+\:\mathrm{sec}^{\mathrm{2}} \:\theta \\ $$

Commented by pieroo last updated on 12/Mar/18

i am grateful

$$\mathrm{i}\:\mathrm{am}\:\mathrm{grateful} \\ $$

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