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

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(B) Given: ${	an ^{ - 1}}x + {	an ^{ - 1}}y + {	an ^{ - 1}}z = \pi $Taking tangent on both sides:${	an ^{ - 1}}x + {	an ^{ - 1}}y = \pi - {	an ^

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(B) Given: ${	an ^{ - 1}}x + {	an ^{ - 1}}y + {	an ^{ - 1}}z = \pi $Taking tangent on both sides:${	an ^{ - 1}}x + {	an ^{ - 1}}y = \pi - {	an ^{ - 1}}z$Applying the formula ${	an ^{ - 1}}A + {	an ^{ - 1}}B = {	an ^{ - 1}}\left( \frac{A+B}{1-AB} ight)$:${	an ^{ - 1}}\left( \frac{x+y}{1-xy} ight) = \pi - {	an ^{ - 1}}z$Taking tangent on both sides:$\frac{x+y}{1-xy} = 	an(\pi - {	an ^{ - 1}}z)$Since $	an(\pi - 	heta) = -	an 	heta$:$\frac{x+y}{1-xy} = -z$$x + y = -z(1 - xy)$$x + y = -z + xyz$$x + y + z = xyz$Dividing both sides by $xyz$:$\frac{x}{xyz} + \frac{y}{xyz} + \frac{z}{xyz} = \frac{xyz}{xyz}$$\frac{1}{yz} + \frac{1}{xz} + \frac{1}{xy} = 1$Thus,$\frac{1}{{xy}} + \frac{1}{{yz}} + \frac{1}{{zx}} = 1$.

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

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