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$\text { If } \tan ^{-1} x+\tan ^{-1} y+\tan ^{-1} z=\frac{\pi}{2}$
then $1-x y-y z-z x$ is equal to
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then $1-x y-y z-z x$ is equal to
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$\tan ^{-1} x+\tan ^{-1} y+\tan ^{-1} z=\frac{\pi}{2}$
$\Rightarrow\left(\tan ^{-1} x+\tan ^{-1} y\right)+\tan ^{-1} z=\frac{\pi}{2}$
$\Rightarrow \tan ^{-1}\left(\frac{x+y}{1-x y}\right)+\tan ^{-1} z=\frac{\pi}{2}$
$\Rightarrow \quad \tan ^{-1}\left[\frac{\frac{(x+y)}{1-x y}+z}{1-z\left(\frac{x+y}{1-x y}\right)}\right]=\frac{\pi}{2}$
$\Rightarrow \frac{(x+y)+z(1-x y)}{(1-x y)-z(x+y)}=\tan \frac{\pi}{2}=\infty$
ie, $\quad(1-x y)-z(x+y)=0$
$\Rightarrow \quad 1-x y-z x-y z=0$
$\Rightarrow\left(\tan ^{-1} x+\tan ^{-1} y\right)+\tan ^{-1} z=\frac{\pi}{2}$
$\Rightarrow \tan ^{-1}\left(\frac{x+y}{1-x y}\right)+\tan ^{-1} z=\frac{\pi}{2}$
$\Rightarrow \quad \tan ^{-1}\left[\frac{\frac{(x+y)}{1-x y}+z}{1-z\left(\frac{x+y}{1-x y}\right)}\right]=\frac{\pi}{2}$
$\Rightarrow \frac{(x+y)+z(1-x y)}{(1-x y)-z(x+y)}=\tan \frac{\pi}{2}=\infty$
ie, $\quad(1-x y)-z(x+y)=0$
$\Rightarrow \quad 1-x y-z x-y z=0$
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