Explanation

(a)
$\because\ \ \ \ \ \ \ y = \frac{10^{x} - 10^{- x}}{10^{x} + 10^{- x}}$
$\Rightarrow \ \ \ \ \ \ \ \frac{y + 1}{y - 1} = \frac{10^{x}}{- 10^{- x}}$
[using componendo and dividendo rule]
$\Rightarrow \ \ \ \ 10^{2x} = \frac{1 + y}{1 - y}$
$\Rightarrow \ \ \ \ \ \ \ \ \ 2x\log_{10}{10 = \log_{10}\left( \frac{1 + y}{1 - y} \right)}$
$\Rightarrow \ \ \ \ \ \ x = \frac{1}{2}\log_{10}\left( \frac{1 + y}{1 - y} \right)$
$\therefore\ \ \ \ \ \ \ f^{- 1}(x) = \frac{1}{2}\log_{10}\left( \frac{1 + x}{1 - x} \right)$