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Question Number 94241 by Ar Brandon last updated on 17/May/20

Prove that arctan(x)+2arctan((√(1+x^2 ))−x)=(π/2)

$$\mathrm{Prove}\:\mathrm{that}\: \\ $$$$\mathrm{arctan}\left(\mathrm{x}\right)+\mathrm{2arctan}\left(\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }−\mathrm{x}\right)=\frac{\pi}{\mathrm{2}} \\ $$

Commented by mathmax by abdo last updated on 17/May/20

let ϕ(x) =arctanx +2 srctan((√(1+x^2 ))−x) ⇒ ϕ^′ (x) =(1/(1+x^2 )) +2 ×((((√(1+x^2 ))−x)^′ )/(1+((√(1+x^2 ))−x)^2 )) =(1/(1+x^2 )) +2×(((x/(√(1+x^2 )))−1)/(1+1+x^2 −2x(√(1+x^2 ))+x^2 )) =(1/(1+x^2 )) +2×((x−(√(1+x^2 )))/((√(1+x^2 ))(2 +2x^2 −2x(√(1+x^2 ))))) =(1/(1+x^2 )) +((x−(√(1+x^2 )))/((√(1+x^2 ))(1+x^2 −x(√(1+x^2 ))))) =(((1+x^2 )(√(1+x^2 ))−x(1+x^2 )+(1+x^2 )x−(1+x^2 )(√(1+x^2 )))/((1+x^2 )(√(1+x^2 ))(1+x^2 −x(√(1+x^2 )))))=0 ⇒ ϕ(x)=c c =ϕ(0) =0 +2arctan(1) =2×(π/4) =(π/2) the equality is proved.

$${let}\:\varphi\left({x}\right)\:={arctanx}\:+\mathrm{2}\:{srctan}\left(\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }−{x}\right)\:\Rightarrow \\ $$$$\varphi^{'} \left({x}\right)\:=\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\:+\mathrm{2}\:×\frac{\left(\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }−{x}\right)^{'} }{\mathrm{1}+\left(\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }−{x}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\:+\mathrm{2}×\frac{\frac{{x}}{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}−\mathrm{1}}{\mathrm{1}+\mathrm{1}+{x}^{\mathrm{2}} −\mathrm{2}{x}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }+{x}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\:+\mathrm{2}×\frac{{x}−\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\left(\mathrm{2}\:+\mathrm{2}{x}^{\mathrm{2}} −\mathrm{2}{x}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\:+\frac{{x}−\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\left(\mathrm{1}+{x}^{\mathrm{2}} −{x}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)} \\ $$$$=\frac{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }−{x}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)+\left(\mathrm{1}+{x}^{\mathrm{2}} \right){x}−\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\left(\mathrm{1}+{x}^{\mathrm{2}} −{x}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)}=\mathrm{0}\:\Rightarrow \\ $$$$\varphi\left({x}\right)={c} \\ $$$${c}\:=\varphi\left(\mathrm{0}\right)\:=\mathrm{0}\:+\mathrm{2}{arctan}\left(\mathrm{1}\right)\:=\mathrm{2}×\frac{\pi}{\mathrm{4}}\:=\frac{\pi}{\mathrm{2}}\:\:{the}\:{equality}\:{is}\:{proved}. \\ $$

Answered by som(math1967) last updated on 17/May/20

tan^(−1) x+2tan^(−1) ((√(1+x^2 ))−x) tan^(−1) x+tan^(−1) ((2((√(1+x^2 ))−x))/(1−((√(1+x^2 ))−x)^2 )) tan^(−1) x+tan^(−1) ((2((√(1+x^2 ))−x))/(1−1−x^2 −x^2 +2x(√(1+x^2 )))) tan^(−1) x+tan^(−1) ((2((√(1+x^2 ))−x))/(2x((√(1+x^2 ))−x))) tan^(−1) x+tan^(−1) (1/x) tan^(−1) x+cot^(−1) x =(π/2)

$$\mathrm{tan}^{−\mathrm{1}} \mathrm{x}+\mathrm{2tan}^{−\mathrm{1}} \left(\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }−\mathrm{x}\right) \\ $$$$\mathrm{tan}^{−\mathrm{1}} \mathrm{x}+\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{2}\left(\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }−\mathrm{x}\right)}{\mathrm{1}−\left(\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }−\mathrm{x}\right)^{\mathrm{2}} } \\ $$$$\mathrm{tan}^{−\mathrm{1}} \mathrm{x}+\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{2}\left(\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }−\mathrm{x}\right)}{\mathrm{1}−\mathrm{1}−\mathrm{x}^{\mathrm{2}} −\mathrm{x}^{\mathrm{2}} +\mathrm{2x}\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }}\: \\ $$$$\mathrm{tan}^{−\mathrm{1}} \mathrm{x}+\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{2}\left(\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }−\mathrm{x}\right)}{\mathrm{2x}\left(\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }−\mathrm{x}\right)} \\ $$$$\mathrm{tan}^{−\mathrm{1}} \mathrm{x}+\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{x}} \\ $$$$\mathrm{tan}^{−\mathrm{1}} \mathrm{x}+\mathrm{cot}^{−\mathrm{1}} \mathrm{x} \\ $$$$=\frac{\pi}{\mathrm{2}} \\ $$

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