How to solve $\tan (a)\cdot \tan (b)>1$

Question

Considering my answer for the question Solve $\cos(a)\cos(b)\cos(a+b) = -1/8$ for $0 < a,b < \pi/2$ I have thought to put

$$x=\tan(a), \quad y=\tan(b)$$

in the inequality

$$\tan (a)\cdot \tan (b)>1 \tag 1$$

I will have

$$xy>1 \tag 2$$

that the $(2)$ it is an equilateral hyperbola when $xy=1$. The crop zone done with Desmos is:

How can I find $a=b=\frac\pi3$?

Using instead Wolfram Alfa I will have many steps https://www.wolframalpha.com/input/?i=%5Ctan+%28a%29%5Ctan+%28b%29%3E1

Is it possible to find with my approach

$$a=b=\frac\pi3\quad ?$$

Answer 1

First of all, we need both $\tan A,\tan B$ of the same sign

If both are $>0,$ which is true in the given range,

$$\tan A>\cot B=\tan(\pi/2-B)$$

As $\tan(x) $ is increasing in the given range of $A,B$ we need $$A>\pi/2-B$$

We find $$-8(1-xy)=(x^2+1)(y^2+1)$$

$$\iff(x-y)^2+(xy-3)^2=0$$

Can you take it from here?

Answer 2

We have $$\tan a > \frac{1}{\tan b} = \cot b = \tan (\frac{\pi}2 - b),$$ hence $a > \frac{\pi}2 - b$ and $a+b > \frac{\pi}2$.