If the sum of all the solutions of ^-1(2x1-x^2)+^-1(1-x^22x)=3, -1

Question

If the sum of all the solutions of $	an^{-1}\left(\dfrac{2x}{1-x^2}ight)+\cot^{-1}\left(\dfrac{1-x^2}{2x}ight)=\dfrac{\pi}{3}$, $-1<x<1,\ x
e0$, is $\alpha-\dfrac{4}{\sqrt3}$, then $\alpha$ is equal to \_\_\_\_\_\_\_.

Answer: 2

Accepted Answer

Case I ($x>0$): For $x>0$, $\cot^{-1}\left(\dfrac{1-x^2}{2x}\right)=\tan^{-1}\left(\dfrac{2x}{1-x^2}\right)$, so the equation becomes

$$2\tan^{-1}\left(\frac{2x}{1-x^2}\right)=\frac{\pi}{3}\ \Rightarrow\ \frac{2x}{1-x^2}=\tan\frac{\pi}{6}=\frac{1}{\sqrt3}\ \Rightarrow\ x^2+2\sqrt3x-1=0\ \Rightarrow\ x=2-\sqrt3.$$

Case II ($x<0$): Here the identity picks up a $+\pi$, giving

$$2\tan^{-1}\left(\frac{2x}{1-x^2}\right)+\pi=\frac{\pi}{3}\ \Rightarrow\ \tan^{-1}\left(\frac{2x}{1-x^2}\right)=-\frac{\pi}{3}\ \Rightarrow\ x=-\frac{1}{\sqrt3}.$$

Sum of solutions $=(2-\sqrt3)+\left(-\dfrac{1}{\sqrt3}\right)=2-\left(\sqrt3+\dfrac{1}{\sqrt3}\right)=2-\dfrac{4}{\sqrt3}$.

Comparing with $\alpha-\dfrac{4}{\sqrt3}$: $\alpha=2$.

Solutions · 1