Question

For the equation ${\cos }^{-1}\left(\frac{x^2-1}{x^2+1}\right)+{\sin }^{-1}\left(\frac{2x}{1+x^2}\right)+{\tan }^{-1}\left(\frac{2x}{x^2-1}\right)=\frac{2\pi }{3},$
which of the following is/are correct?

• A. The number of solution is $2$
• B. It has atleast one irrational solution.
• C. The sum of solution is positive.
• D. The sum of solutions is negative.

Answer 1

Answer: (B), (C)

The correct options are
B It has atleast one irrational solution.
C The sum of solution is positive.

${\cos }^{-1}\left(\frac{x^2-1}{x^2+1}\right)+{\sin }^{-1}\left(\frac{2x}{1+x^2}\right)+{\tan }^{-1}\left(\frac{2x}{x^2-1}\right)=\frac{2\pi }{3}\pi -{\cos }^{-1}\left(\frac{1-x^2}{x^2+1}\right)+{\sin }^{-1}\left(\frac{2x}{1+x^2}\right)-{\tan }^{-1}\left(\frac{2x}{1-x^2}\right)=\frac{2\pi }{3}$
Putting $x=\tan \theta$, we have
${\cos }^{-1}\left(\cos 2\theta \right)+{\tan }^{-1}\left(\tan 2\theta \right)-{\sin }^{-1}\left(\sin 2\theta \right)=\frac{\pi }{3}\theta \in (0,\frac{\pi }{4})\Rightarrow 2\theta =\frac{\pi }{3}\Rightarrow x=\frac{1}{\sqrt{3}}\theta \in (\frac{\pi }{4},\frac{\pi }{2})\Rightarrow 2\theta +(2\theta -\pi )-(\pi -2\theta )=\frac{\pi }{3}\Rightarrow \theta =\frac{7\pi }{18}\frac{7\pi }{18}\Rightarrow x=\tan \left(\frac{7\pi }{18}\right)=\tan \left({70}^{\circ }\right)\theta \in (\frac{-\pi }{4},0)\Rightarrow -2\theta +2\theta -2\theta =\frac{\pi }{3}\Rightarrow x=-\frac{1}{\sqrt{3}}\left(\text{irrational}\right)$
So, sum = $\tan \left({70}^{\circ }\right)$