<s>[INST] Approximation of $\pi$ using Brahmagupta's Identity Brahmagupta, an ancient Indian Mathematician, gave an pretty efficient algorithm for finding integer solutions to the famous Pell's Equation, far before Fermat propounded this before the European mathematicians' community. 

Brahmagupta's Identity:
If $(x_1,y_1)$ is a solution to $Dx^2+m=y^2$ and $(x_2,y_2)$ is a solution to $Dx^2+n=y^2$, then $(x_1y_2\pm x_2y_1,y_1y_2\pm D x_1x_2)$ is a solution to the equation $Dx^2+mn=y^2$.

Famous mathematician André Weil denoted this more efficiently by $$(x_1,y_1;m)\oplus (x_2,y_2;n)=(x_1y_2\pm x_2y_1,y_1y_2\pm D x_1x_2;mn)$$ 
One can easily prove this by writing $m=y_1^2-Dx_1^2$ and $n=y_2^2-Dx_2^2$, and then multiplying them $$mn=(y_1^2-Dx_1^2)(y_2^2-Dx_2^2)=(y_1y_2\pm D x_1x_2)^2-D(x_1y_2\pm x_2y_1)^2$$ and also notice that it is a group. A $600 AD$ mathematician is solving problems using Group Theory! 
I am giving an example, integer solutions to $83x^2+1=y^2$.
We know that, $$83\times 1^2-2=9^2. $$
So, we here get $$(1,9;-2)\oplus (1,9;-2)=(18,81+83;4)=(18,164;4).$$
So, we get the equation, $$\begin{align}
&83(18)^2+4=(164)^2\\
\implies &83\left(\frac {18}2\right)^2+1=\left(\frac {164}2\right)^2\\
\implies &83\times 9^2+1=82^2
\end{align}$$
Now, we have, if $Da^2+1=b^2$, then, $$\frac ba-\sqrt D=\frac {b-\sqrt D a}a=\frac {b^2-Da^2}{a(b+\sqrt D a)}=\frac 1{a(b+\sqrt D a)} $$

So, for sufficiently large $(a,b)$, $\frac ba$ is a good approximation for $\sqrt D$.

One, can verify by finding solutions to $2a^2+1=b^2$, i.e. $(2,3),(12,17),\dots$. So $$\color{red}{\sqrt 2\approx \frac 32,\frac {17}{12},\frac {577}{408}}.$$

So, I was trying to approximate $\sqrt \pi$ or $e$ or $\pi$ using this identity, but could not came to result. I am trying my way, but you folks please help me sharing your idea.

As, $\pi$ is irrational, we can't have $\pi =D$, so, I used $3\lt \pi\lt 4$, so $\sqrt 3\lt \sqrt \pi\lt 2$, so, $1\lt \sqrt \pi \lt 2$,
But feeling some difficulty.
 [/INST] The procedure for calculating the square root of the number can be used to calculate the number $\pi$ to arbitrary precision. You can use the ratio $$\tan\dfrac x2 = \dfrac{\tan x} {\sqrt{\tan^2 x+1} + 1}$$ with initial data $\tan{\dfrac{\pi}4} = 1$.
After $n$ iterations you will have $\tan\dfrac{\pi}{2^{n+2}}$ and then can use the formula $$\pi=\lim\limits_{n\to\infty} 2^{n+2}\tan\dfrac{\pi}{2^{n+2}}$$ or  Maclaurin series for the arctangent function:
$$\pi=2^{n+2}\left(\tan\dfrac{\pi}{2^{n+2}}-\dfrac13\left(\tan\dfrac{\pi}{2^{n+2}}\right)^3+\dots\right)$$
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