As easy as pie

$A = \frac 4{1+\frac {1^2}{3+\frac {2^2}{5+\frac {3^2}{7+\frac {4^2}{9+\ddots}}}}} = \boxed \pi$ . See reference (I can't prove this.)

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From $\cos x = 2\cos^2 \frac x2 -1$ $\implies \cos \frac x2 = \sqrt{\frac {1+\cos x}2}$ $\implies \cos \frac x4 = \sqrt{\frac {1+\cos \frac x2}2} = \sqrt{\frac 12 + \frac 12 \sqrt{\frac {1+\cos x}2}}$ $\implies \cos \frac x8 = \sqrt{\frac 12 + \frac 12 \sqrt{\frac 12 + \frac 12 \sqrt{\frac {1+\cos x}2}}}$ , $\cdots$

For $x = \frac \pi 4$ , $\cos \frac \pi 4 = \frac {\sqrt 2}2$ , $\cos \frac \pi 8 = \frac {\sqrt {2+\sqrt 2}}2$ , $\cos \frac \pi{16} = \frac {\sqrt {2+\sqrt{2+\sqrt 2}}}2$ , $\cos \frac \pi{32} = \frac {\sqrt {2+\sqrt{2+\sqrt {2+\sqrt 2}}}}2$ , $\cdots$

Therefore,

$\begin{aligned} \frac 2B & = \frac {\sqrt 2}2 \times \frac {\sqrt{2+\sqrt 2}}2 \times \frac {\sqrt{2+\sqrt {2+\sqrt 2}}}2 \times \cdots \\ & = \cos \frac \pi 4 \cos \frac \pi 8 \cos \frac \pi{16} \cos \frac \pi{32} \cdots \\ & = \lim_{n \to \infty} \prod_{k=2}^n \cos \frac \pi{2^k} \\ & = \lim_{n \to \infty} \prod_{k=2}^{\color{#D61F06}n-1} \cos \frac \pi{2^k} \times \frac {\color{#3D99F6}\sin \frac \pi{2^n}\color{#D61F06}\cos \frac \pi{2^n}}{\color{#3D99F6}\sin \frac \pi{2^n}} = \lim_{n \to \infty} \prod_{k=2}^{n-1} \cos \frac \pi{2^k} \times \frac {\sin \frac \pi{2^{\color{#3D99F6}n-1}}}{{\color{#3D99F6}2}\sin \frac \pi{2^n}} \\ & = \lim_{n \to \infty} \prod_{k=2}^{\color{#D61F06}n-2} \cos \frac \pi{2^k} \times \frac {\color{#3D99F6}\sin \frac \pi{2^{n-1}}\color{#D61F06}\cos \frac \pi{2^{n-1}}}{2\sin \frac \pi{2^n}} = \lim_{n \to \infty} \prod_{k=2}^{n-2} \cos \frac \pi{2^k} \times \frac {\sin \frac \pi{2^{\color{#3D99F6}n-2}}}{{\color{#3D99F6}4}\sin \frac \pi{2^n}} \\ & = \lim_{n \to \infty} \frac {\sin \frac \pi2}{2^{n-1}\sin \frac \pi{2^n}} = \lim_{n \to \infty} \frac 2{2^n\sin \frac \pi{2^n}} \\ \implies B & = \lim_{n \to \infty} 2^n\sin \frac \pi{2^n} = \lim_{n \to \infty} \pi \cdot \frac {\sin \frac \pi{2^n}}{\frac \pi{2^n}} = \boxed \pi \end{aligned}$

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By Maclaurin series , we have:

$\begin{aligned} \tan^{-1}x & = x - \frac {x^3}3 + \frac {x^5}5 - \frac {x^7}7 + \frac {x^9}9 - \cdots & \small \color{#3D99F6} \text{Putting }x = 1 \\ \implies \frac \pi 4 & = 1 - \frac 13 + \frac 15 - \frac 17 + \frac 19 - \cdots = \frac C4 \\ \implies C & = \boxed \pi \end{aligned}$

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$\begin{aligned} D & = 3 \color{#3D99F6}+ \frac 4{2 \times 3 \times 4} \color{#D61F06}- \frac 4{4 \times 5 \times 6} \color{#3D99F6}+ \frac 4{6 \times 7 \times 8} \color{#D61F06}- \frac 4{8 \times 9 \times 10} \color{#3D99F6}+ \cdots & \small \color{#3D99F6} \text{By partial fraction decomposition} \\ & = 3 \color{#3D99F6}+ \frac 22 - \frac 43 + \frac 24 \color{#D61F06}- \frac 24 + \frac 45 - \frac 26 \color{#3D99F6}+ \frac 26 - \frac 47 + \frac 28 \color{#D61F06}- \frac 28 + \frac 49 - \frac 2{10} \color{#3D99F6}+ \cdots \\ & = 4 - \frac 43 + \frac 45 - \frac 47 + \frac 49 - \cdots = C = \boxed \pi \end{aligned}$

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Therefore, $A=B=C=D=\boxed \pi$ . Answer: All of these choices.