The possible values of $x$, if $\tan^{-1} x>\cot^{-1}x$

Question

What are the possible values of $x$, if $\tan^{-1}x >\cot^{-1}x$?

We have $\tan^{-1}x >\cot^{-1}x\implies \tan^{-1}x -\cot^{-1}x>0 \implies \tan^{-1} x-\tan^{-1}1/x>0\implies \tan^{-1}\dfrac{x-1/x}{1-x.1/x}>0$.
What can I do now?

Answer 1

HINT:

$$\cot^{-1}x=\frac\pi2-\tan^{-1}x\implies \tan^{-1}x-\cot^{-1}x=2\tan^{-1}x-\frac\pi2>0$$

Answer 2

Given $\displaystyle \tan^{-1}(x)>\cot^{-1}(x)\;\forall\; x\in\mathbb{R}.$

So $\displaystyle \tan^{-1}(x)>\frac{\pi}{2}-\tan^{-1}(x)$

So $\displaystyle \tan^{-1}(x)>\frac{\pi}{4}\Rightarrow x>1$

So we get $x\in(1\;,\infty).$

Answer 3

A plot might help to guide the derivation.

Recall that for $x\ge0$

$$\cot^{-1}(x)=\frac{\pi}{2}-\tan^{-1}(x)$$

and hence

$$\tan^{-1}(x)>\frac{\pi}{2}-\tan^{-1}(x)$$

$$ \tan^{-1}(x)>\frac{\pi}{4}\implies x>1$$

and for $x<0$

$$\cot^{-1}(x)=-\frac{\pi}{2}-\tan^{-1}(x)$$

hence

$$\tan^{-1}(x)>-\frac{\pi}{2}-\tan^{-1}(x)$$

$$ \tan^{-1}(x)>-\frac{\pi}{4}\implies x>-1$$

therefore the solution is

$$x\in (-1,0)\cup(-1,\infty)$$