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Solve for x:$\tan^{-1}(x+1)+\tan^{-1}x+\tan^{-1}(x-1)=\tan^{-1}3.$

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  • \( tan^{-1}x + tan^{-1}y = tan^{-1}\large\frac{x + y}{1 - xy}, xy < 1 \)
Using the above formula and taking x+1 in the place of x and x-1 in the place of y we get
\( tan^{-1}(x+1)+tan^{-1}(x-1)=tan^{-1}\large\frac{x+1+x-1}{1-(x^2-1)}\)
\(= tan^{-1} \bigg(\large \frac{2x}{1-x^2+1} \bigg) = tan^{-1} \bigg(\large\frac{2x}{2-x^2} \bigg) \)
And using the same formula we get
\(tan^{-1}3-tan^{-1}x=tan^{-1}\large\frac{3-x}{1+3x} \)
We can write the given equation as
\( tan^{-1}(x+1)+tan^{-1}(x-1)=tan^{-1}3-tan^{-1}x\)
substituting the values on either side we get
\( \Rightarrow\:tan^{-1}\large\frac{2x}{2-x^2}= tan^{-1}\large\frac{3-x}{1+3x}\)
\( \Rightarrow 6x^2+2x=6+x^3-3x^2-2x\)
\(\Rightarrow\: x^3-9x^2-4x+6=0\)
\(\Rightarrow\: (x+1)(x^2-10x+6)=0\)
\( x=-1 \: or \: x=\large\frac{10 \pm \sqrt{76}}{2}=5 \pm \sqrt {19} \)


answered Feb 19, 2013 by thanvigandhi_1
edited Mar 19, 2013 by thanvigandhi_1

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