We have $\mathrm{f}\left(\mathrm{x}\right)=\cot \mathrm{x}-\sqrt{2}\mathrm{cosec}\mathrm{x}=\frac{\cos \mathrm{x}-\sqrt{2}}{\sin \mathrm{x}}$
 So,  ${\mathrm{f}}^{\mathrm{\prime }}\left(\mathrm{x}\right)=\frac{\sin \mathrm{x}(-\sin \mathrm{x})-(\cos \mathrm{x}-\sqrt{2})\cos \mathrm{x}}{{\sin }^2\mathrm{x}}=\frac{-1+\sqrt{2}\cos \mathrm{x}}{{\sin }^2\mathrm{x}}$

∴  ${\mathrm{f}}^{\mathrm{\prime }}\left(\mathrm{x}\right)=0\Rightarrow \cos \mathrm{x}=\frac{1}{\sqrt{2}}\Rightarrow \mathrm{x}=\frac{\mathrm{\pi }}{4}\in (0,\mathrm{\pi })$

$\therefore \mathrm{f}\left(\mathrm{x}\right)\uparrow \text{ on }\left(0,\frac{\mathrm{\pi }}{4}\right)\text{ and }\mathrm{f}\left(\mathrm{x}\right)\downarrow \text{ on }\left(\frac{\mathrm{\pi }}{4},\mathrm{\pi }\right)$
    Also, $\underset{\mathrm{x}\to 0^{+}}{\mathrm{Lim}}\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{x}\to 0^{+}}{\mathrm{Lim}}\left(\frac{\mathrm{cosx}-\sqrt{2}}{\mathrm{sinx}}\right)\to –\mathrm{\infty }\text{ and }\underset{\mathrm{x}\to {\mathrm{\pi }}^{-}}{\mathrm{Lim}}\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{x}\to {\mathrm{\pi }}^{-}}{\mathrm{Lim}}\left(\frac{\mathrm{cosx}-\sqrt{2}}{\mathrm{sinx}}\right)\to –\mathrm{\infty }$
and f(x) is also continuous on (0, π) 
Clearly f(x) < 0 ∀ x ∈ (0, π)
∴    At x =$\frac{\mathrm{\pi }}{4}$   (a local maximum point), so f(x) takes its absolute maximum value also at x  = $\frac{\mathrm{\pi }}{4}$.
Hence, absolute maximum value of  $\mathrm{f}\left(\mathrm{x}=\frac{\mathrm{\pi }}{4}\right)=1-\sqrt{2}\left(\sqrt{2}\right)=1-2=-1$ Ans.