If ${\tan }^{-1}\mathrm{x}+{\tan }^{-1}\mathrm{y}+{\tan }^{-1}\mathrm{z}=\mathrm{\pi }/2$ then $1-\mathrm{xy}-\mathrm{yz}-\mathrm{zx}=$

$\text{ Let }{\tan }^{-1}\mathrm{x}=\alpha$

$$\begin{gathered} \begin{array}{l}\Rightarrow \tan \alpha =x\\ {\tan }^{-1}y=\beta \\ \Rightarrow \tan \beta =y\\ {\tan }^{-1}z=\gamma \\ \Rightarrow \tan \gamma =z\\ \Rightarrow \alpha +\beta +\gamma =\frac{\pi }{2}\\ \tan (\alpha +\beta +\gamma )=\frac{\tan \alpha +\tan \beta +\tan \gamma -\tan \alpha \tan \beta \tan \gamma }{1-(\tan \alpha \tan \beta +\tan \beta \tan \gamma +\tan \gamma \tan \alpha )} \\ =\tan \frac{\pi }{2}=not defined \\ \end{array} \end{gathered}$$

$\begin{array}{r}\Rightarrow \tan \alpha \tan \beta +\tan \beta \tan \gamma +\tan \gamma \tan \alpha =1\\ \Rightarrow \mathrm{xy}+\mathrm{yz}+\mathrm{zx}=1\end{array}$