Range of $y=\cot^{-1}x\cot^{-1}(-x)$ 
Find the range of $y = \cot^{-1}x\cot^{-1}(-x)$ over the whole real line.

Attempt: Let $\cot^{-1}x=\theta$. Then $y=\theta(\pi-\theta)$; since $\theta \in (0,\pi) $, $y>0$. Then $\theta ^2 -\pi \theta+y =0 $. For $\theta$ to be real, $y\le \dfrac{\pi^2}4$.
Thus $0<y\le \dfrac{\pi^2}4$.
Though this is the right answer I am highly unsure about my step "for $\theta$ to be real" because $\theta \in (0,\pi)$. Is that step correct? Why or why not? 
Also, any alternative methods to solve this problem? 
 A: I don't fully understand the logic behind your method by discriminant, as an alternative from here $$\theta ^2 -\pi \theta+y =0 \implies y=\pi \theta-\theta ^2=\theta(\pi-\theta) \quad \theta \in (0,\pi)$$
which is a concave down parabola such that $y(0)=0$, $y(\pi)=0$ and is maximum for $\theta=\frac{\pi}{2}$ that is $y\left(\frac{\pi}{2}\right)=\dfrac{\pi^2}{4}$ thus the range is $0<y\le \dfrac{\pi^2}{4}$.
A: The step you have inquired about is correct; we may find the values of $y$ for which $\theta^2-\pi\theta+y=0$ has (at least one) real solution through its discriminant $\pi^2-4\cdot1\cdot y$, which has to be non-negative. This gives your inequality $y\le\frac{\pi^2}4$.
Another way to solve the problem is to use the identity $\cot^{-1}x=\frac\pi2-\tan^{-1}x$ to rewrite the given function as
$$y=\left(\frac\pi2-\tan^{-1}x\right)\left(\frac\pi2-\tan^{-1}(-x)\right)$$
Since $\tan^{-1}x$ is an odd function:
$$y=\left(\frac\pi2-\tan^{-1}x\right)\left(\frac\pi2+\tan^{-1}x\right)=\frac{\pi^2}4-(\tan^{-1}x)^2$$
Since $\tan^{-1}x$ has range $(-\pi/2,\pi/2)$, $(\tan^{-1}x)^2$ has range $[0,\pi^2/4)$, giving the range of $y$ as $(0,\pi^2/4]$.
