Given, $\cos h2x=199$

Now,$\cos h2x=\frac{e^{2x}+e^{-2x}}{2}$

$\Rightarrow \frac{e^{2x}+e^{-2x}}{2}=199$

$\Rightarrow e^{2x}+e^{-2x}=398$

Now, $\cot hx=\frac{e^x+e^{-x}}{e^x-e^{-x}}$

${\left(a+b\right)}^2=a^2+b^2+2ab$

$\Rightarrow {\left(e^x+e^{-x}\right)}^2=e^{2x}+e^{-2x}+2e^x{e}^{-x}$

$\Rightarrow {\left(e^x+e^{-x}\right)}^2=398+2=400$

$\Rightarrow \left(e^x+e^{-x}\right)=\sqrt{400}=20$

${\left(a-b\right)}^2=a^2+b^2-2ab$

$\Rightarrow {\left(e^x-e^{-x}\right)}^2=e^{2x}+e^{-2x}-2e^x{e}^{-x}$

$\Rightarrow {\left(e^x-e^{-x}\right)}^2=398-2=396$

$\Rightarrow \left(e^x-e^{-x}\right)=\sqrt{396}=6\sqrt{11}$

Therefore, $\cot hx=\frac{e^x+e^{-x}}{e^x-e^{-x}}=\frac{20}{6\sqrt{11}}=\frac{10}{3\sqrt{11}}$