If $0<x<\pi /2\text{ and }\sin (2\sin x)=\cos (2\cos x)$ then $\tan x+\cot x=\frac{a}{{\pi }^C-b}\text{ where }a+b+c=$

$\begin{array}{l}\sin (2\sin x)=\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\sin (\pi /2-2\cos x)\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}2\sin x=\pi /2-2\cos x\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\sin x+\cos x=\pi /4\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1+\sin 2x={\pi }^2/16\end{array}$

$\begin{array}{l}\tan x+\cot x=\frac{2}{\sin 2x}=\frac{32}{{\pi }^2-16}\\ a=32, b=16, c=2\end{array}$