If the range of $y=\left({\cot }^{-1}x\right)\left({\cot }^{-1}(-x)\right)$ is $(K,\frac{L}{4}]$  Then the value of $K+\frac{16L}{{\pi }^2}$ is ____________.

$y=\left({\cot }^{-1}x\right)\left({\cot }^{-1}\left(x\right)\right)=\left({\cot }^{-1}x\right)(\pi -{\cot }^{-1}\left(x\right))$

Now, ${\cot }^{-1}\left(x\right)$ and $(\pi -{\cot }^{-1}\left(x\right))>0$ 

Using $A.M\ge G.M.,$ we get $\frac{{\cot }^{-1}x+(\pi -{\cot }^{-1}x)}{2}\ge \sqrt{\left({\cot }^{-1}x\right)(\pi -{\cot }^{-1}x)}$

$\Rightarrow 0<{\cot }^{-1}\left(x\right)(\pi -{\cot }^{-1}\left(x\right))\le {\left(\frac{{\cot }^{-1}x+(\pi -{\cot }^{-1}x)}{2}\right)}^2=\frac{{\pi }^2}{4}\Rightarrow 0<y<\frac{{\pi }^2}{4}$