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Question Number 127402 by bemath last updated on 29/Dec/20

Compute x if arctan x+arctan 1 = 2(arctan x−arctan (1/3))

$$\:{Compute}\:{x}\:{if}\:\mathrm{arctan}\:{x}+\mathrm{arctan}\:\mathrm{1}\:=\:\mathrm{2}\left(\mathrm{arctan}\:{x}−\mathrm{arctan}\:\frac{\mathrm{1}}{\mathrm{3}}\right)\: \\ $$

Answered by liberty last updated on 29/Dec/20

⇔ arctan x+arctan 1=2arctan x−2arctan (1/3) ⇔arctan x = arctan 1+2arctan (1/3) tan (arctan x) = tan ((π/4)+2arctan (1/3)) x = ((1+tan (2arctan (1/3)))/(1−tan (2arctan (1/3)))) (∗) tan (2arctan (1/3))= ((2tan (arctan (1/3)))/(1−tan^2 (arctan (1/3)))) = ((2/3)/(1−1/9)) = (3/4) ⇔ x = ((1+(3/4))/(1−(3/4))) = 7.

$$\Leftrightarrow\:\mathrm{arctan}\:{x}+\mathrm{arctan}\:\mathrm{1}=\mathrm{2arctan}\:{x}−\mathrm{2arctan}\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\Leftrightarrow\mathrm{arctan}\:{x}\:=\:\mathrm{arctan}\:\mathrm{1}+\mathrm{2arctan}\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\:\:\:\:\:\mathrm{tan}\:\left(\mathrm{arctan}\:{x}\right)\:=\:\mathrm{tan}\:\left(\frac{\pi}{\mathrm{4}}+\mathrm{2arctan}\:\frac{\mathrm{1}}{\mathrm{3}}\right) \\ $$$$\:\:\:\:\:{x}\:=\:\frac{\mathrm{1}+\mathrm{tan}\:\left(\mathrm{2arctan}\:\frac{\mathrm{1}}{\mathrm{3}}\right)}{\mathrm{1}−\mathrm{tan}\:\left(\mathrm{2arctan}\:\frac{\mathrm{1}}{\mathrm{3}}\right)}\: \\ $$$$\left(\ast\right)\:\mathrm{tan}\:\left(\mathrm{2arctan}\:\frac{\mathrm{1}}{\mathrm{3}}\right)=\:\frac{\mathrm{2tan}\:\left(\mathrm{arctan}\:\frac{\mathrm{1}}{\mathrm{3}}\right)}{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \left(\mathrm{arctan}\:\frac{\mathrm{1}}{\mathrm{3}}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{2}/\mathrm{3}}{\mathrm{1}−\mathrm{1}/\mathrm{9}}\:=\:\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\:\Leftrightarrow\:{x}\:=\:\frac{\mathrm{1}+\frac{\mathrm{3}}{\mathrm{4}}}{\mathrm{1}−\frac{\mathrm{3}}{\mathrm{4}}}\:=\:\mathrm{7}. \\ $$

Commented by bemath last updated on 29/Dec/20

Danke Meister. Ihre Antwort ist wunderschön

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