The inverse of the function $y=\frac{{10}^x-{10}^{-x}}{{10}^x+{10}^{-x}}$  is

Since, $y=\frac{{10}^{2x}-1}{{10}^{2\times }+1}$

$\Rightarrow y^{{10}^{2x}}+y={10}^{2x}-1$

$\Rightarrow {10}^{2x}=\frac{1+y}{1-y}$

$\therefore$   $2x{\log }_{10}10={\log }_{10}\frac{1+y}{1-y}$

$\Rightarrow x=\frac{1}{2}{\log }_{10}\left(\frac{1+y}{1-y}\right)$

Hence, the inverse of y is $\frac{1}{2}{\log }_{10}\left(\frac{1+y}{1-y}\right)$ .