I have some problem with the theory of mathematical induction.
Suppose I have a recursive formulae; I assumed a function and I proved it by induction that it satisfying the recursion.
But does it imply that the function is the only one satisfying the recursion?
[Updated.]
Here we are assuming $a_n= \cot{\frac{\pi}{2^{n+1}}}$. But is it the only one satisfying the relation?
Your question lacks sufficient details to answer specifically. In general, induction does not guarantee uniqueness, but there are cases when you can prove that the solution is unique.
Which one your specific case is depends on the recursion and what you managed to prove exactly.
For example, consider $a_n = a_{n-1}$ with $a_0=k$ for any $k \in \mathbb{R}$. If you use induction to prove $a_n \equiv k \forall n \in \mathbb{N}$ then it is easy to see this solution is unique.
On the other hand, consider $a_n - 3a_{n-1} + 2a_{n-2}=0$ with no initial conditions specified. Then you have 2 solution families $a_n \equiv k$ and $a_n = c \cdot 2^n$ which will both satisfy the recurrence, so if you prove $k$ satisfies it, or if you prove $c \cdot 2^n$ satisfies it, neither will guarantee uniqueness.
UPDATE
For your specific problem, since the initial conditions are specified, there is only one sequence that satisfies the recurrence. This is easy to see by repeatedly plugging in the previous value to get the next one, starting with the initial condition.
How you express that sequence may be different, i.e. $a_n = (-1)^n$ can also be expressed as $a_n = \sin(n \pi)$, but it cannot change what the actual values of the sequence are. Hence, if there is any one representation which makes it convenient to take the limit, you could use it.