If tan ^{-1} x+tan ^{-1} y+tan ^{-1} z=frac{p | vedclass

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(B) Given that $	an ^{-1} x+	an ^{-1} y+	an ^{-1} z=\frac{\pi}{2}$.We know the formula $	an ^{-1} x+	an ^{-1} y = 	an ^{-1} \left( \frac{x+y}{1-

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(B) Given that $	an ^{-1} x+	an ^{-1} y+	an ^{-1} z=\frac{\pi}{2}$.We know the formula $	an ^{-1} x+	an ^{-1} y = 	an ^{-1} \left( \frac{x+y}{1-xy} ight)$.So, $	an ^{-1} \left( \frac{x+y}{1-xy} ight) + 	an ^{-1} z = \frac{\pi}{2}$.Taking $	an$ on both sides, we use $	an(A+B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$.Let $A = 	an ^{-1} \left( \frac{x+y}{1-xy} ight)$ and $B = 	an ^{-1} z$.Then $	an(A+B) = 	an \left( \frac{\pi}{2} ight) = \infty$.For $	an(A+B)$ to be $\infty$, the denominator must be $0$.Thus, $1 - 	an A 	an B = 0$.$1 - \left( \frac{x+y}{1-xy} ight) z = 0$.$1 - \frac{xz + yz}{1-xy} = 0$.$1 - xy - xz - yz = 0$.Therefore, $1 - xy - yz - zx = 0$.

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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