A geometry problem by Aakhyat Singh

$\begin{cases} \csc A - \sin A = a^3 & \implies a^3 = \dfrac 1{\sin A} - \sin A = \dfrac {1-\sin^2 A}{\sin A} = \dfrac {\cos^2 A}{\sin A} \\ \sec A - \cos A = b^3 & \implies b^3 = \dfrac 1{\cos A} - \cos A = \dfrac {1-\cos^2 A}{\cos A} = \dfrac {\sin^2 A}{\cos A} \end{cases}$

$\begin{aligned} \implies \frac {a^3}{b^3} & = \frac {\cos^3 A}{\sin^3 A} \implies \frac ab = \frac {\cos A}{\sin A} \end{aligned}$

$\begin{aligned} \implies a^3b^3 & = \cos A \sin A \end{aligned}$

Then, we have:

$\begin{aligned} \frac {\cos A}{\sin A} + \frac {\sin A}{\cos A} & = \frac {\cos^2 A + \sin^2 A}{\sin A \cos A} \\ \frac ab + \frac ba & = \frac 1{a^3b^3} \\ \frac {a^2+b^2}{ab} & = \frac 1{a^3b^3} \\ a^2+b^2 & = \frac 1{a^2b^2} \end{aligned}$

$\quad \quad \implies \boxed{a^2b^2 = \dfrac 1{a^2+b^2}}$