Trigo-fying

$A=cos {\frac{\pi}{5}}cos {\frac{2 \pi}{5}}=\frac{2sin {\frac{\pi}{5}}cos {\frac{\pi}{5}}cos {\frac{2 \pi}{5}}}{2 sin {\frac{\pi}{5}}}= \frac{sin {\frac{2 \pi}{5}} cos {\frac{2 \pi}{5}}}{2 sin {\frac{\pi}{5}}} =\frac{2sin {\frac{2 \pi}{5}} cos {\frac{2 \pi}{5}}}{4 sin {\frac{\pi}{5}}}=\frac{sin {\frac{4 \pi}{5}}}{4sin {\frac{\pi}{5}}}=\frac{sin {\frac{\pi}{5}}}{4sin {\frac{\pi}{5}}}=\frac{1}{4}$

$B=cos {\frac{\pi}{5}}-cos {\frac{2 \pi}{5}}=2 sin {\frac{3 \pi}{10}} sin {\frac{\pi}{10}}=2 cos {\frac{\pi}{5}}cos {\frac{2 \pi}{5}}=2A=2 \times\frac{1}{4}=\frac{1}{2}$

Therefore,

$\frac{1}{A}+\frac{1}{B}=4+2=\boxed{6}$

We note that $\displaystyle \sum_{k=0}^{n-1} \cos \frac {2k+1}{2n+1} \pi = \dfrac 12$ (see proof ). Then we have,

$\begin{aligned} \cos \frac \pi 5 {\color{#3D99F6}+ \cos \frac {3 \pi}5} & = \frac 12 & \small \color{#3D99F6} \text{Note that }\cos (\pi - x) = - \cos x \\ \cos \frac \pi 5 {\color{#3D99F6}- \cos \frac {2 \pi}5} & = \frac 12 \\ \implies B & = \frac 12 \end{aligned}$

$\begin{aligned} A & = \cos \frac \pi 5 \cos \frac {2 \pi}5 \\ & = \frac 12 \left(\cos \left(\frac {2 \pi}5 - \frac \pi 5 \right) + \cos \left(\frac {2 \pi}5 + \frac \pi 5 \right) \right) \\ & = \frac 12 \left(\cos \frac \pi 5 + \cos \frac {3\pi}5 \right) \\ & = \frac 12 \left(\frac 12 \right) \\ & = \frac 14 \end{aligned}$

$\implies \dfrac 1A + \dfrac 1B = 4 + 2 = \boxed{6}$