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

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

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$xy + yz + zx - 1 = 0$

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(D) Given that $	an ^{-1}x + 	an ^{-1}y + 	an ^{-1}z = \frac{\pi }{2}.$Using the formula $	an ^{-1}x + 	an ^{-1}y + 	an ^{-1}z = 	an ^{-1}\left( \frac{x + y + z - xyz}{1 - xy - yz - zx} ight),$$	an ^{-1}\left( \frac{x + y + z - xyz}{1 - xy - yz - zx} ight) = \frac{\pi }{2}.$Taking tangent on both sides:$\frac{x + y + z - xyz}{1 - xy - yz - zx} = 	an \left( \frac{\pi }{2} ight) = \infty.$For the expression to be undefined (approaching $\infty$),the denominator must be zero:$1 - xy - yz - zx = 0.$Rearranging the terms,we get:$xy + yz + zx = 1$ or $xy + yz + zx - 1 = 0.$Thus,option $(d)$ is correct.

If $\tan ^{-1}x + \tan ^{-1}y + \tan ^{-1}z = \frac{\pi }{2},$ then

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