Limiting a Secant Product

Let

$\text{P} = \displaystyle \prod_{r=1}^{n} \left(\sec\left(\frac{x}{2^{r-1}}\right) + 1 \right)$

Note that,

$\displaystyle \sec x +1 = \dfrac{1+\cos x}{\cos x}$

$\displaystyle = \dfrac{2\cos^2 \left(\frac{x}{2}\right) \cdot \tan x}{\sin x}$

$\displaystyle = \dfrac{\tan x}{\tan \left(\frac{x}{2}\right)}$

$\displaystyle \implies \text{P} = \prod_{r=1}^{n} \dfrac{\tan \left(\frac{x}{2^{r-1}} \right)}{\tan \left(\frac{x}{2^r}\right)} \ (*)$

Clearly, $(*)$ telescopes. Evaluating it, we have,

$\text{P} = \displaystyle \dfrac{\tan x}{\tan \left( \frac{x}{2^n} \right)}$

$\implies \displaystyle f(x) = \lim_{n \to \infty} 2^{-n} \cdot \text{P} = \lim_{n \to \infty} 2^{-n} \cdot \dfrac{\tan x}{\tan \left( \frac{x}{2^n} \right)} = \dfrac{\tan x}{x}$

$\implies \displaystyle f' \left( \frac{\pi}{4} \right) = \dfrac{8}{\pi} - \dfrac{16}{{\pi}^2}$

$\displaystyle \therefore A + B = \boxed{24}$