If ${\tan }^3\theta +{\cot }^3\theta =8{\mathrm{cosec}}^{3}2\theta +12$ then A=

$\frac{{\sin }^3\theta }{{\cos }^3\theta }+\frac{{\cos }^3\theta }{{\sin }^3\theta }=\frac{8}{{\sin }^{3}2\theta }+12$

$\begin{array}{l}\Rightarrow {\sin }^6\theta +{\cos }^6\theta =1+12{\sin }^3\theta {\cos }^3\theta \\ \Rightarrow \left({\sin }^2\theta +{\cos }^2\theta \right)\left({\sin }^4\theta -{\sin }^2\theta {\cos }^2\theta +{\cos }^4\theta \right)\end{array}$

  $=1+\frac{3}{2}{\sin }^{3}2\theta$

$\begin{array}{l}\Rightarrow 1-3{\sin }^2\theta {\cos }^2\theta =1+\frac{3}{2}{\sin }^{3}2\theta \\ \Rightarrow -\frac{1}{4}{\sin }^{2}2\theta =\frac{1}{2}{\sin }^{3}2\theta \Rightarrow {\sin }^{2}2\theta (1+2\sin 2\theta )=0\end{array}$

      $\sin 2\theta \ne 0,\text{ since }\theta \ne 0,\frac{\pi }{2}\therefore \sin 2\theta =-\frac{1}{2}=\sin \left(\frac{-\pi }{6}\right)$    

$\Rightarrow 2\theta =n\pi +(-1{)}^n\left(\frac{-\pi }{6}\right)\Rightarrow \theta =n\pi -\frac{\pi }{12},n\pi -\frac{5\pi }{12}$