Mistakes give rise to Problems- 10

This thing is true $\color{#D61F06}{\textbf{if and only if}}$ $a=b$

(Because though $\cos(a)=\cos(-a)$ , if we take $b=-a$ then LHS=1, RHS= -1.

Thus there are $\boxed{20}$ pairs, as $a=b \neq 0$ , so they can be integers $-10$ to $10$ except $0$ .

namely, they are $(-10,-10),(-9,-9),....,(-1,-1),(1,1),(2,2),....,(9,9),(10,10)$

I'd like to add info from Aditya's solution...

Yes... the equation is true if and only if a = b with the help of symmetric property of cosine: cos(-x) = cos (x)...

Suppose we contradict the fact that there are other pairs aside from a = b that satisfies the equation.

Consider the increasing and decreasing behavior of cosine as x tends to 0 (maybe using a bit of Calculus here), the cosine is increasing as x goes to 0. This means the function is increasing as x goes to 0. So that means if a > b, then cos(a) < cos b (satisfying the condition) this means that a/b > 1 and cos (a) / cos (b) < 1, contradiction.

Hence, there are 20 ordered pairs.

I created a brute-force algorithm for c++ to check each pair $(a, b)$ such that $-10 \leq a, b \leq 10$ and $b \neq 0$ .

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12

#define PI 3.14159265

int main () {
    int count = 0;
    for(int a = -10; a <= 10; a++) {
        for(int b = -10; b <= 10; b++) {
            if(!b) continue; //skip b = 0 
            if(cos(a*PI/180) / cos(b*PI/180) == a/b) count++;
        }
    }
    cout << "Result is " << count << endl;
}

The result is $\boxed{20}$ .