$\frac{e^{4x}+e^{-4x}+14}{4{\left(e^x-e^{-x}\right)}^2}=$

Given,

$\frac{e^{4x}+e^{-4x}+14}{4{\left(e^x-e^{-x}\right)}^2}$

Now solving option one by one to match the given expression,

Now we know that,

The formula of $\sin hx=\frac{e^x-e^{-x}}{2}$ and $\cot hx=\frac{e^x+e^{-x}}{e^x-e^{-x}}$

Now solving $\cot h^{2}x+\sin h^{2}x=\left(\frac{e^{2x}+e^{-2x}+2}{e^{2x}+e^{-2x}-2}\right)+\left(\frac{e^{2x}+e^{-2x}-2}{4}\right)$

$\Rightarrow \cot h^{2}x+\sin h^{2}x=\left(\frac{4e^{2x}+4e^{-2x}+8+\left(e^{2x}+e^{-2x}-2\right)\left(e^{2x}+e^{-2x}-2\right)}{4\left(e^{2x}+e^{-2x}-2\right)}\right)$

$\Rightarrow \cot h^{2}x+\sin h^{2}x=\left(\frac{4e^{2x}+4e^{-2x}+8+{\left(e^{2x}+e^{-2x}-2\right)}^2}{4{\left(e^x-e^{-x}\right)}^2}\right)$

$\Rightarrow \cot h^{2}x+\sin h^{2}x=\frac{4e^{2x}+4e^{-2x}+8+e^{4x}+e^{-4x}+4+2-4e^{2x}-4e^{-2x}}{4{\left(e^x-e^{-x}\right)}^2}$

$\Rightarrow \cot h^{2}x+\sin h^{2}x=\frac{e^{4x}+e^{-4x}+14}{4{\left(e^x-e^{-x}\right)}^2}$