${\tan }^{-1}\left(\mathrm{a}\right) +\hspace{0.17em}{\tan }^{-1}\left(\mathrm{b}\right) +\hspace{0.17em}{\tan }^{-1}\left(\mathrm{c}\right) = \mathrm{\pi }$

$\therefore \hspace{0.17em}{\tan }^{-1}\left(\mathrm{b}\right) +\hspace{0.17em}{\tan }^{-1}\left(\mathrm{c}\right) = \mathrm{\pi } - {\tan }^{-1}\left(\mathrm{a}\right)$

$\therefore {\tan }^{-1}\left(\frac{\mathrm{b} + \mathrm{c}}{1 - \mathrm{bc}}\right) = \mathrm{\pi } - {\tan }^{-1}\left(\mathrm{a}\right)$

$\therefore \frac{\mathrm{b} + \mathrm{c}}{1 - \mathrm{bc}} = \tan \left(\mathrm{\pi } - {\tan }^{-1}\left(\mathrm{a}\right)\right)$

$\therefore \frac{\mathrm{b} + \mathrm{c}}{1 - \mathrm{bc}} = - \tan \left( {\tan }^{-1}\left(\mathrm{a}\right)\right)$

$\therefore \frac{\mathrm{b} + \mathrm{c}}{1 - \mathrm{bc}} = - \mathrm{a}$

$\therefore \mathrm{b} + \mathrm{c} = - \mathrm{a} + \mathrm{abc}$

$\therefore \mathrm{a} + \mathrm{b}\hspace{0.17em}+ \mathrm{c} = \mathrm{abc}$

Given,

$\tan \left({\cos }^{-1}\left(\mathrm{x}\right)\right) = \frac{2}{\sqrt{5}}$

We have, ${\cos }^{-1}\left(\mathrm{x}\right) = {\tan }^{-1}\left(\frac{\sqrt{1 - {\mathrm{x}}^2}}{\mathrm{x}}\right)$

$\therefore \tan \left({\tan }^{-1}\left(\frac{\sqrt{1 - {\mathrm{x}}^{2 }}}{\mathrm{x}}\right)\right) = \frac{2}{\sqrt{5}}$

$\therefore \frac{\sqrt{1 - {\mathrm{x}}^2}}{\mathrm{x}}= \frac{2}{\sqrt{5}}$

Squaring on both sides,

$\frac{1 - {\mathrm{x}}^2}{\mathrm{x}} = \frac{4}{5}$

5 - 5x² = 4x²

$$\begin{gathered} \therefore 9{\mathrm{x}}^2 = 5 \\ \therefore \mathrm{x} = \frac{\sqrt{5}}{3} \end{gathered}$$

Given, $\mathrm{a} * \mathrm{b} = 2\mathrm{a} + \mathrm{b}$

$\therefore$ (2*3)*4 = $\left(2 \times 2 + 3\right) * 4$

                 = 7 * 4

                 = $2 \times 7 + 4$

                 = 14 + 4

                 = 18

Given equation is, $3{\tan }^{-1}\left(\mathrm{x}\right) + {\cot }^{-1}\left(\mathrm{x}\right) = \mathrm{\pi }$

$\therefore 2{\tan }^{-1}\left(\mathrm{x}\right) + {\tan }^{-1}\left(\mathrm{x}\right) + {\cot }^{-1}\left(\mathrm{x}\right) = \mathrm{\pi }$

$\therefore 2{\tan }^{-1}\left(\mathrm{x}\right) + \frac{\mathrm{\pi }}{2} = \mathrm{\pi }$

$\therefore 2{\tan }^{-1}\left(\mathrm{x}\right) = \mathrm{\pi } - \frac{\mathrm{\pi }}{2}$

$\therefore {\tan }^{-1}\left(\mathrm{x}\right) = \frac{\mathrm{\pi }}{2} - \frac{\mathrm{\pi }}{4}$

$\therefore \mathrm{x} =\tan \left(\frac{\mathrm{\pi }}{2} - \frac{\mathrm{\pi }}{4}\right)$

$\therefore \mathrm{x} = 1$

let, y = $\sqrt{2\mathrm{x} - 3}$

$\therefore {\mathrm{y}}^2 = 2\mathrm{x} - 3$

$\therefore \mathrm{x} = \frac{{\mathrm{y}}^2 + 3}{2}$

$\therefore {\mathrm{f}}^{-1}\left(\mathrm{x}\right) = \frac{{\mathrm{x}}^2 +\hspace{0.17em}3}{2}$

Now, L.H.S. = ${\mathrm{fof}}^{- 1 }\left(\mathrm{x}\right) = \mathrm{f}\left[{\mathrm{f}}^{- 1}\left(\mathrm{x}\right)\right]$

                  = $\sqrt{2{\mathrm{f}}^{- 1}\left(\mathrm{x}\right) - 3}$

                  = $\sqrt{2\left(\frac{{\mathrm{x}}^2 +\hspace{0.17em}3}{2}\right) - 3}$

                  = $\sqrt{{\mathrm{x}}^2}$

                  = x

  $\therefore {\mathrm{fof}}^{- 1}\left(\mathrm{x}\right) = \mathrm{x}$

Hence proved.