Does this property holds for inverse too? (part- Vl)

First note that the function $f(x) = \tan^{-1}(x)\cot^{-1}(x)$ is a continuous, even function that is differentiable on $\mathbb{R} - \{0\}.$ We also note that $f(0) = 0.$ Now using the product rule we find that

$f'(x) = \dfrac{\cot^{-1}(x)}{1 + x^{2}} - \dfrac{\tan^{-1}(x)}{1 + x^{2}},$

which equals $0$ when $\tan^{-1}(x) = \cot^{-1}(x).$ This occurs only when $x = 1,$ at which point $f(1) = \dfrac{\pi}{4}*\dfrac{\pi}{4} = \dfrac{\pi^{2}}{16} \lt 1.$ Since $f(1) \gt f(0)$ we know that the critical point obtained represents a maximum, and thus we can conclude that $f(x) \lt 1$ $\forall x \in \mathbb{R},$ and so the given equation has no real solutions.

Using the property $tan^{-1}x + cot^{-1}x=\pi/2$ The equation becomes $(tan^{-1}x)^{2}-\pi/2 (tan^{-1}x)+1$ Clearly if we substitute $tan^{-1}x=y$ The quadratic equation so formed has no solution.