Trigonometric equations: cotangent

Question

If I have

$cot(x-a)=cot(x-b)$

Where x is in radians and equal on both the sides and not equal to $0$ or $π$ Also for a and b, they are not equal to $0$ or $π$

Does the above equality mean $a=b$? If not then how do we even find the value of a and b? Do we need any more conditions?

There are no singularity points of cotangent between points $(x-a)$ and $(x-b)$

Answer 1

Is $a=b$ ?

Not necessarily. Because $\cot$ is periodic with a period of $n\pi , n\in \Bbb Z$.

So, $\cot(x-a) = \cot(x-b) = \cot(n\pi+x-b) \implies \color{purple}{a =m\pi +b} , m =-n \in \Bbb Z$

Answer 2

Hint: Use that $$\cot(x)-\cot(y)=-\csc (x) \csc (y) \sin (x-y)$$

Answer 3

$\cot$ is a periodic function that repeats its value after ever $π$ interval i.e $\cot(X) = \cot (nπ + X)$.