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

## 1 Answer

Toolbox:
• $$tan^{-1}x+tan^{-1}y=tan^{-1}\bigg( \large\frac{x+y}{1-xy} \bigg)$$
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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