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

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

Solution:

To find: value of $x$
Formula Used: $\tan ^{-1} \mathrm{x}+\tan ^{-1} \mathrm{y}=\tan ^{-1}\left(\frac{\mathrm{x}+y}{1-\mathrm{xy}}\right)$ where $\mathrm{xy}<1$
Given: $\tan ^{-1}(x+1)+\tan ^{-1}(x-1)=\tan ^{-1} \frac{a}{31}$
$\begin{array}{l} \text { LHS }=\tan ^{-1}\left(\frac{x+1+x-1}{1-((x+1) x(x-1))}\right) \\ =\tan ^{-1} \frac{2 x}{1-\left(x^{2}-x+x-1\right)} \\ =\tan ^{-1} \frac{2 x}{2-x^{2}} \end{array}$
Therefore, $\tan ^{-1} \frac{2 x}{2-x^{2}}=\tan ^{-1} \frac{8}{31}$
Taking tangent on both sides, we get
$\begin{array}{l} \frac{2 x}{2-x^{2}}=\frac{8}{31} \\ \Rightarrow 62 x=16-8 x^{2} \\ \Rightarrow 8 x^{2}+62 x-16=0 \\ \Rightarrow 4 x^{2}+31 x-8=0 \\ \Rightarrow 4 x^{2}+32 x-x-8=0 \\ \Rightarrow 4 x \times(x+8)-1 \times(x+8)=0 \\ \Rightarrow(4 x-1) \times(x+8)=0 \\ \Rightarrow x=\frac{1}{4} \text { or } x=-8 \end{array}$
Therefore, $x=\frac{1}{4}$ or $x=-8$ are the required values of $x$ .