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Question Number 114988 by ZiYangLee last updated on 22/Sep/20

Prove tan^2 x=sin^2 x+sec^2 x

$$\mathrm{Prove}\:\mathrm{tan}^{\mathrm{2}} {x}=\mathrm{sin}^{\mathrm{2}} {x}+\mathrm{sec}^{\mathrm{2}} {x} \\ $$

Answered by Olaf last updated on 22/Sep/20

It′s false sir. tan^2 x ≠ sin^2 x+sec^2 x

$$\mathrm{It}'\mathrm{s}\:\mathrm{false}\:\mathrm{sir}. \\ $$$$\mathrm{tan}^{\mathrm{2}} {x}\:\neq\:\mathrm{sin}^{\mathrm{2}} {x}+\mathrm{sec}^{\mathrm{2}} {x} \\ $$

Commented by ZiYangLee last updated on 22/Sep/20

Ok sir tks

$$\mathrm{Ok}\:\mathrm{sir}\:\mathrm{tks} \\ $$

Answered by Olaf last updated on 22/Sep/20

But tan^2 x = ((sin^2 x)/(cos^2 x)) = sin^2 x((1/(cosx)))^2 = sin^2 x×sec^2 x It′s × not +

$$\mathrm{But}\:\mathrm{tan}^{\mathrm{2}} {x}\:=\:\frac{\mathrm{sin}^{\mathrm{2}} {x}}{\mathrm{cos}^{\mathrm{2}} {x}}\:=\:\mathrm{sin}^{\mathrm{2}} {x}\left(\frac{\mathrm{1}}{\mathrm{cos}{x}}\right)^{\mathrm{2}} =\:\mathrm{sin}^{\mathrm{2}} {x}×\mathrm{sec}^{\mathrm{2}} {x} \\ $$$$\mathrm{It}'\mathrm{s}\:×\:\mathrm{not}\:+ \\ $$$$ \\ $$

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