If a B is an acute angle and (1-tan a) (1-tan b) = 2, then a + B=

If a B is an acute angle and (1-tan a) (1-tan b) = 2, then a + B=

It is estimated that only Chinese people will come up with this type of topic
This paper mainly uses the formula: Tan (a + b) = (Tana + tanb) / (1 - Tana * tanb) 1
If the observation is inconvenient, let's assume Tan (a) = a, Tan (b) = B
Then ① can be rewritten as: Tan (a + b) = (a + b) / (1 - AB) ③
Now let's look at your question, (1-tan a) (1-tan b) = 2
Let's rewrite it
It is known that: (1-A) (1-B) = 2.5
Finding the value of "anti cotangent function" of Tan (a + b)
Expansion 3: 1 - A - B + AB = 2
A + B = 1 - AB ⑥
(a + b) / (1-ab) = 1 = Tan (a + b) isn't this just the right part of ③
So: a + B = 45 °± K × 180 ° a, B for acute angle, can only be k = 0
The conclusion is: a + B = 45 degree
In the morning of dandelion, why do I divide (1-ab) by (1-ab) in formula 6?