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Question Number 128755 by bramlexs22 last updated on 10/Jan/21

Solve tan^(−1) (x−1)+tan^(−1) x +tan^(−1) (x+1) = tan^(−1) 3x

$$\mathrm{Solve}\:\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{x}−\mathrm{1}\right)+\mathrm{tan}^{−\mathrm{1}} \mathrm{x}\:+\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{x}+\mathrm{1}\right)\:=\:\mathrm{tan}^{−\mathrm{1}} \mathrm{3x} \\ $$

Answered by liberty last updated on 10/Jan/21

” tan^(−1) (x−1)+tan^(−1) (x+1)=tan^(−1) (3x)−tan^(−1) (x)   ((2x)/(1−(x^2 −1))) = ((2x)/(1+3x^2 )) ; ((2x)/(2−x^2 )) = ((2x)/(1+3x^2 ))  ⇔ 2x {1+3x^2 −2+x^2  } = 0  ⇔2x (4x^2 −1)=0 → { ((x=0)),((x=± (1/2))) :}

$$''\:\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{x}−\mathrm{1}\right)+\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{x}+\mathrm{1}\right)=\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{3x}\right)−\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{x}\right) \\ $$$$\:\frac{\mathrm{2x}}{\mathrm{1}−\left(\mathrm{x}^{\mathrm{2}} −\mathrm{1}\right)}\:=\:\frac{\mathrm{2x}}{\mathrm{1}+\mathrm{3x}^{\mathrm{2}} }\:;\:\frac{\mathrm{2x}}{\mathrm{2}−\mathrm{x}^{\mathrm{2}} }\:=\:\frac{\mathrm{2x}}{\mathrm{1}+\mathrm{3x}^{\mathrm{2}} } \\ $$$$\Leftrightarrow\:\mathrm{2x}\:\left\{\mathrm{1}+\mathrm{3x}^{\mathrm{2}} −\mathrm{2}+\mathrm{x}^{\mathrm{2}} \:\right\}\:=\:\mathrm{0} \\ $$$$\Leftrightarrow\mathrm{2x}\:\left(\mathrm{4x}^{\mathrm{2}} −\mathrm{1}\right)=\mathrm{0}\:\rightarrow\begin{cases}{\mathrm{x}=\mathrm{0}}\\{\mathrm{x}=\pm\:\frac{\mathrm{1}}{\mathrm{2}}}\end{cases} \\ $$

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