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

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## 1 Answer

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• $tan^{-1}x+tan^{-1}y=tan^{-1}\bigg( \large\frac{x+y}{1-xy} \bigg)$
By taking x+1 in the place of x and x-1 in the place of y we get
$\frac{x+y}{1-xy}=\frac{(x+1)+(x-1)}{1-(x+1).(x-1)}=\frac{2x}{1-(x^2-1)}=\frac{2x}{2-x^2}$
Substituting in the above formula of $tan^{-1}x+tan^{-1}y$ we get
$tan^{-1}(x+1)+tan^{-1}(x-1)=tan^{-1}\frac{2x}{2-x^2}=tan^{-1}\frac{8}{31}$
$\Rightarrow\:tan^{-1} \bigg(\large \frac{2x}{2-x^2} \bigg) = tan^{-1}\large\frac{8}{31}$
$\Rightarrow \large\frac{2x}{2-x^2}=\large\frac{8}{31}$
$\Rightarrow\:62x=8(2-x^2)$
$\Rightarrow\:31x=4(2-x^2)$
$\Rightarrow\:4x^2+31x-8=0$
$\Rightarrow\:4x^2+32x-x-8=0$
$\Rightarrow\:4x(x+8)-(x+8)=0$
$\Rightarrow\:(4x-1)(x+8)=0$
Solving which we get$x=-8\:\:or\:x=\large \frac{1}{4}$
But $x = -8$ is not in principal interval so $x= \large\frac{1}{4}$
answered Feb 28, 2013
edited Mar 19, 2013

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