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

Toolbox:
• $tan^{-1}x+tan^{-1}y=tan^{-1}\large\frac{x+y}{1-xy}, xy< 1$
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
$tan^{-1}(x+1)+tan^{-1}(x-1)=tan^{-1}\bigg(\frac{(x+1)+(x-1)}{1-(x+1).(x-1)}\bigg)$
$=tan^{-1}\bigg(\frac{2x}{1-(x^2-1)}\bigg)=tan^{-1}\frac{2x}{2-x^2}$
Substituting the value in the given eqn., we have
$tan^{-1}(x+1)+tan^{-1}(x-1)= tan^{-1}\bigg( \large\frac{2x}{2-x^2} \bigg)=tan^{-1}\large\frac{8}{31}$
$\Rightarrow\: \large\frac{2x}{2-x^2}=\large\frac{8}{31}$
$\Rightarrow\large\:31x=8-4x^2$
$\Rightarrow\:\large\:4x^2+31x-8=0$
$\Rightarrow\:x=\large\frac{1}{4},8$
$x=8$ doesn't satisfying
$Ans : \: so \: x=\large\frac{1}{4}$
edited Mar 22, 2013