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Solve \( \tan^{-1}(x+1)+tan^{-1}(x-1)=tan^{-1}\large\frac{8}{31} \).

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  • \( tan^{-1}x+tan^{-1}y=tan^{-1}\large\frac{x+y}{1-xy}, xy< 1\)
By taking x+1 in the place of x and x-1 in the place of y in the
formula of \(tan^{-1}x+tan^{-1}y\) we have
Substituting the value in the given eqn., we have
\(tan^{-1}(x+1)+tan^{-1}(x-1)= tan^{-1}\bigg( \large\frac{2x}{2-x^2} \bigg)=tan^{-1}\large\frac{8}{31}\)
\(\Rightarrow\: \large\frac{2x}{2-x^2}=\large\frac{8}{31}\)
\( \Rightarrow\:x=\large\frac{1}{4},8\)
\( x=8\) doesn't satisfying
\( Ans : \: so \: x=\large\frac{1}{4}\)
answered Feb 28, 2013 by thanvigandhi_1
edited Mar 22, 2013 by rvidyagovindarajan_1
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