The sum of possible values of x for ^-1(x+1)+^-1(1x-1)=^-1831 is

Question

The sum of possible values of $x$ for $	an^{-1}(x+1)+\cot^{-1}\left(\dfrac{1}{x-1}ight)=	an^{-1}\dfrac{8}{31}$ is

Accepted Answer

$\cot^{-1}\left(\dfrac{1}{x-1}\right)=\tan^{-1}(x-1)$, so

$$\tan^{-1}(x+1)+\tan^{-1}(x-1)=\tan^{-1}\frac{8}{31}\ \Rightarrow\ \tan^{-1}\left(\frac{2x}{2-x^2}\right)=\tan^{-1}\frac{8}{31}.$$

$$\Rightarrow\frac{2x}{2-x^2}=\frac{8}{31}\ \Rightarrow\ 62x=16-8x^2\ \Rightarrow\ 8x^2+62x-16=0\ \Rightarrow\ 2(x+8)(4x-1)=0\ \Rightarrow\ x=-8,\ \frac14.$$

Checking $x=\dfrac14$: $\tan^{-1}\dfrac54+\cot^{-1}\left(-\dfrac43\right)$, where each term individually exceeds $\dfrac{\pi}{2}$, so the LHS exceeds $\dfrac{\pi}{2}$ while the RHS $\tan^{-1}\dfrac{8}{31}<\dfrac{\pi}{2}$ --- this root is spurious (introduced by the tangent-addition identity) and is rejected.

So $x=-8$ is the only solution; the sum of possible values is $-8=-\dfrac{32}{4}$.

Solutions · 1