Magnetic field at the centre of regular polygon of  'n' sides which is formed by wire, which carries current I and side of polygon is 'a':-

(1) $\frac{\mathrm{n}{\mu }_0\mathrm{I}}{\mathrm{\pi a}}\sin \left(\frac{\mathrm{\pi }}{\mathrm{n}}\right)\cot \left(\frac{\mathrm{\pi }}{\mathrm{n}}\right)$                        (2) $\frac{\mathrm{n}{\mu }_0\mathrm{I}}{\mathrm{\pi a}}\cos \left(\frac{\mathrm{\pi }}{\mathrm{n}}\right)\tan \left(\frac{\mathrm{\pi }}{\mathrm{n}}\right)$
(3) $\frac{\mathrm{n}{\mu }_0\mathrm{I}}{\mathrm{\pi a}}\sin \left(\frac{\mathrm{\pi }}{\mathrm{n}}\right)\tan \left(\frac{\mathrm{\pi }}{\mathrm{n}}\right)$                        (4) $\frac{\mathrm{n}{\mu }_0\mathrm{I}}{\mathrm{\pi a}}$

Dear Student ,

Let us take an n-sided regular polygon as shown in the figure.

From the figure, we can see that:
$$\begin{gathered} 2x=2a\sin \left(\frac{\pi }{n}\right) \\ \Rightarrow x=a\sin \frac{\pi }{n} ...\left(1\right) \\ y=a\cos \frac{\pi }{n} ...\left(2\right) \end{gathered}$$
Using Biot Savart law the magnetic field at the center due to one of the side,
$$\begin{gathered} B=\frac{{\mu }_{0}I}{4\pi y}\left(\sin {\varphi }_1+\sin {\varphi }_2\right) \\ =\frac{{\mu }_{0}I}{4\pi y}\left(\sin \frac{\pi }{n}+\sin \frac{\pi }{n}\right) \\ =\frac{{\mu }_{0}I}{4\pi y}\left(2\sin \frac{\pi }{n}\right) \\ =\frac{{\mu }_{0}I}{2\pi y}\sin \frac{\pi }{n} \end{gathered}$$
Therefore, the magnetic field at the center due to 'n' number of sides:

$$\begin{gathered} B=\frac{n{\mu }_{0}I}{2\pi y}\sin \frac{\pi }{n} \\ \Rightarrow B=\frac{n{\mu }_{0}I}{2\pi a\cos \frac{\pi }{n}}\sin \frac{\pi }{n} (\text{using eq}.(2\left)\right) ...\left(3\right) \\ \Rightarrow B=\frac{n{\mu }_{0}I}{2\pi a}\tan \frac{\pi }{n} ...\left(4\right) \end{gathered}$$
Let us take the case when 'n' tends to infinity:
Taking the limit in equation (3):
$$\begin{gathered} B=\underset{n\to \infty }{Lt}\frac{n{\mu }_{0}I}{2\pi a\cos \frac{\pi }{n}}\sin \frac{\pi }{n} \\ Let t=\frac{\pi }{n} \\ \Rightarrow n=\frac{\pi }{t} \\ \text{As} n\to \infty , t\to 0 \\ \Rightarrow B=\underset{t\to 0}{Lt}\frac{{\mu }_{0}I}{2\pi a}\frac{\pi }{t}\frac{\sin t}{\cos t}=\frac{{\mu }_{0}I}{2a}\underset{t\to 0}{Lt}\frac{{\displaystyle \frac{\sin t}{t}}}{\cos t} \\ \text{As} \underset{t\to 0}{Lt}\frac{\sin t}{t}=1 and \underset{t\to 0}{Lt}\cos t=1 \\ \text{Therefore,} \\ B=\frac{{\mu }_{0}I}{2a} \end{gathered}$$
Thus, the formula reduces to the magnetic field at the centre due to the circular loop.
Regards