1tan -1 (x-1) + tan-1(x+1) = tan-13x -tan-1x,
tan-1(x-1+x +1)/1 - (x-1)(x+y)=tan-1(3x-x)/1+3x2 ,
2x /2-x2 = 2x/1 +3x2,
x =0 or when x≠0 ,
4x2 = 1,
x =±1/2
$Solve the equation\\\\ $tan^{-1}(x-1)+tan^{-1}(x)+tan^{-1}(x+1)=tan^{-1}(3x)$
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3 Answers
30$tan^{-1}(x-1)+tan^{-1}(x)+tan^{-1}(x+1)=tan^{-1}(3x)$
\Rightarrow\tan^{-1} (x-1 ) + \tan^{-1}(x+1) = \tan^{-1}(3x)-\tan^{-1}(x)
\Rightarrow \frac{tan(\tan^{-1} (x-1 )) + tan(\tan^{-1}(x+1))}{1-tan(\tan^{-1} (x-1 ))tan(\tan^{-1}(x+1))} = \frac{tan(\tan^{-1}(3x))-tan(\tan^{-1}(x)) }{1+tan(\tan^{-1}(3x))tan(\tan^{-1}(x))}
\Rightarrow \frac{x-1+x+1}{1-(x-1)(x+1)}=\frac{3x-x}{1+(3x)(x)}
\Rightarrow \frac{2x}{2-x^2}=\frac{2x}{1+3x^2}
\Rightarrow\boxed {x=0}
or when x≠0,
\Rightarrow 2-x^2=1+3x^2
\Rightarrow 4x^2=1
\Rightarrow \boxed{x=\pm 1/2}
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