The sum of possible values of x for tan ^{-1} | vedclass

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(B) Given equation: $	an ^{-1}(x+1)+\cot ^{-1}\left(\frac{1}{x-1}ight)=	an ^{-1}\left(\frac{8}{31}ight)$.Using the property $\cot ^{-1}(u) = 	a

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$-\frac{31}{4}$

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(B) Given equation: $	an ^{-1}(x+1)+\cot ^{-1}\left(\frac{1}{x-1}ight)=	an ^{-1}\left(\frac{8}{31}ight)$.Using the property $\cot ^{-1}(u) = 	an ^{-1}\left(\frac{1}{u}ight)$ for $u > 0$,we have $\cot ^{-1}\left(\frac{1}{x-1}ight) = 	an ^{-1}(x-1)$ if $x-1 > 0$,i.e.,$x > 1$.If $x > 1$,then $	an ^{-1}(x+1) + 	an ^{-1}(x-1) = 	an ^{-1}\left(\frac{8}{31}ight)$.Applying $	an(A+B)$ formula: $\frac{(x+1)+(x-1)}{1-(x+1)(x-1)} = \frac{8}{31} \Rightarrow \frac{2x}{1-(x^2-1)} = \frac{8}{31} \Rightarrow \frac{2x}{2-x^2} = \frac{8}{31}$.$62x = 16 - 8x^2 \Rightarrow 8x^2 + 62x - 16 = 0 \Rightarrow 4x^2 + 31x - 8 = 0$.Solving for $x$: $(4x-1)(x+8) = 0$,so $x = \frac{1}{4}$ or $x = -8$. Both values are $\leq 1$,so the assumption $x > 1$ is invalid.If $x Then $	an ^{-1}(x+1) + \pi + 	an ^{-1}(x-1) = 	an ^{-1}\left(\frac{8}{31}ight)$.$	an ^{-1}\left(\frac{2x}{2-x^2}ight) = 	an ^{-1}\left(\frac{8}{31}ight) - \pi$.Taking $	an$ on both sides: $\frac{2x}{2-x^2} = 	an(	an ^{-1}(\frac{8}{31}) - \pi) = \frac{8}{31}$.This leads to $4x^2 + 31x - 8 = 0$,giving $x = \frac{1}{4}$ or $x = -8$. Since $x  \pi/2$. Thus $x = 1/4$ is rejected.For $x = -8$: $	an^{-1}(-7) + \cot^{-1}(-1/9) = -1.42 + 1.68 = 0.26 = 	an^{-1}(8/31)$.Thus,the only solution is $x = -8$.

The sum of possible values of $x$ for $\tan ^{-1}(x+1)+\cot ^{-1}\left(\frac{1}{x-1}\right)=\tan ^{-1}\left(\frac{8}{31}\right)$ is

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