Inverse trigonometric inequalities

Question

If $$ \frac{-\pi}{2}<\tan^{-1}x,\tan^{-1}y<\frac{\pi}{2}\quad\&\quad\frac{\pi}{2}<\tan^{-1}x+\tan^{-1}y<\pi\quad\&\quad xy>1 $$

How do I mathematically prove that $x>0$ and $y>0$ ?

My Attempt:

I can see that if $\tan^{-1}x$ or $\tan^{-1}y$ is less than zero $\tan^{-1}x+\tan^{-1}y\ngtr\frac{\pi}{2}$. But how do i prove it mathematically ?

Answer 1

Note that

• $\frac{-\pi}{2}<\tan^{-1}x \quad\tan^{-1}y<\frac{\pi}{2}$ is always true

Since $ xy>1$ we have that x and y are both positive or both negative.

For x and y both negative

$$\tan^{-1}x+\tan^{-1}y<0$$

and the given inequality is not verified.

Whereas for x and y both positive

$$xy>1 \implies x>\frac1y \implies \tan^{-1}x>\tan^{-1}\frac1y=\frac{\pi}2-\tan^{-1}y \implies \tan^{-1}x+\tan^{-1}y>\frac{\pi}2$$

Answer 2

We have $xy>1$ so either both $x$ and $y$ are positive or negative. Suppose the latter. Then $\tan^{-1}x+\tan^{-1}y<0$ so by contradiction $x,y>0$.