Always the complementary ones!

We note that $A+B=\dfrac \pi 2$ , $\implies \tan B = \tan (\frac \pi 2 - A) = \cot A = \dfrac 1{\tan A}$ . Then we have:

$\begin{aligned} \tan A + \tan B + \tan^2 A + \tan^2 B + \tan^3 A + \tan^3 B & = 70 \\ \tan A + \frac 1{\tan A} + \tan^2 A + \frac 1{\tan^2 A} + \tan^3 A + \frac 1{\tan^3 A} & = 70 & \small \color{#3D99F6}{\text{Let }x = \tan A} \\ x + \frac 1x + x^2 + \frac 1{x^2} + x^3 + \frac 1{x^3} & = 70 \\ x + \frac 1x + x^2 + \color{#3D99F6}{2} + \frac 1{x^2} - \color{#D61F06}{2} + x^3 + \color{#3D99F6}{3x + \frac 3x} + \frac 1{x^3} - \color{#D61F06}{3x + \frac 3x} & = 70 \\ \left(x + \frac 1x\right) + \left(x + \frac 1x\right)^2 - \color{#D61F06}{2} + \left(x + \frac 1x\right)^3 - \color{#D61F06}{3\left(x + \frac 1x\right)} & = 70 \\ \left(x + \frac 1x\right)^3 + \left(x + \frac 1x\right)^2 - 2\left(x + \frac 1x\right) - 72 & = 0 \\ \implies x + \frac 1x & = 4 \\ x^2 - 4x + 1 & = 0 \\ x = \tan A & = 2 \pm \sqrt 3 \\ \implies A, B & = \boxed{\dfrac {5 \pi}{12}, \ \dfrac {\pi}{12}} \end{aligned}$