Question

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

Answer 1

Here $x+x^{2}+x^{3}+x^{4}+\ldots \infty=\frac{x}{1-x} ;|x| < 1$
and $-6+6 x-6 x^{2}+\ldots \infty=\frac{-6}{1+x} ;|x| < 1$
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} $
$\Rightarrow x^{2}+x=-6+6 x$
$\Rightarrow x^{2}-5 x+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.