Question

For $x\in (-1,1)$, the number of solutions of the equation ${\tan }^{-1}\left(x+x^2+x^3+x^4+\dots \infty \right)+{\cot }^{-1}(-6+6x-6x^2+\dots$ $\infty )=\frac{\pi }{2}$, is

Answer 1

Answer: 0

Here $x+x^2+x^3+x^4+\dots \infty =\frac{x}{1-x};|x|<1$
and $-6+6x-6x^2+\dots \infty =\frac{-6}{1+x};|x|<1$
So, that ${\tan }^{-1}\left(\frac{x}{1-x}\right)+{\cot }^{-1}\left(\frac{-6}{1+x}\right)=\frac{\pi }{2}$
$\Rightarrow \frac{x}{1-x}=\frac{-6}{1+x}$
$\Rightarrow x^2+x=-6+6x$
$\Rightarrow x^2-5x+6=0$
$\Rightarrow (x-2)(x-3)=0$
$\Rightarrow x=2,3$
But $-1<x<1$ (Given)
So, we conclude that there is no solution.