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Question Number 182842 by malwan last updated on 15/Dec/22

prove that  sec(tan^(−1) x)=(√(x^2 +1))

$${prove}\:{that} \\ $$$${sec}\left({tan}^{−\mathrm{1}} {x}\right)=\sqrt{{x}^{\mathrm{2}} +\mathrm{1}} \\ $$

Answered by cortano1 last updated on 15/Dec/22

 let x = tan t    (√(x^2 +1)) = (√(tan^2 t+1)) = (√(sec^2 t)) = sec t  ⇒sec (tan^(−1) (tan t))= sec t

$$\:{let}\:{x}\:=\:\mathrm{tan}\:{t}\: \\ $$$$\:\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\:=\:\sqrt{\mathrm{tan}\:^{\mathrm{2}} {t}+\mathrm{1}}\:=\:\sqrt{\mathrm{sec}\:^{\mathrm{2}} {t}}\:=\:\mathrm{sec}\:{t} \\ $$$$\Rightarrow\mathrm{sec}\:\left(\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{tan}\:{t}\right)\right)=\:\mathrm{sec}\:{t}\: \\ $$

Commented by malwan last updated on 15/Dec/22

thank you sir

$${thank}\:{you}\:{sir} \\ $$

Answered by BaliramKumar last updated on 15/Dec/22

LHS = sec(tan^(−1) x)  LHS = (√(sec^2 (tan^(−1) x)))  LHS = (√(tan^2 (tan^(−1) x)+1))  LHS = (√(x^2 +1)) = RHS

$${LHS}\:=\:{sec}\left({tan}^{−\mathrm{1}} {x}\right) \\ $$$${LHS}\:=\:\sqrt{{sec}^{\mathrm{2}} \left({tan}^{−\mathrm{1}} {x}\right)} \\ $$$${LHS}\:=\:\sqrt{{tan}^{\mathrm{2}} \left({tan}^{−\mathrm{1}} {x}\right)+\mathrm{1}} \\ $$$${LHS}\:=\:\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\:=\:{RHS} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\: \\ $$

Commented by malwan last updated on 15/Dec/22

thank you so much

$${thank}\:{you}\:{so}\:{much} \\ $$

Answered by manxsol last updated on 16/Dec/22

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