If tan ^{-1}left(frac{x}{2}right)+tan ^{-1}le | vedclass

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(D) Given that $	an ^{-1}\left(\frac{x}{2}ight)+	an ^{-1}\left(\frac{y}{2}ight)+	an ^{-1}\left(\frac{z}{2}ight)=\frac{\pi}{2}$.Let $A = 	an

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$4$

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(D) Given that $	an ^{-1}\left(\frac{x}{2}ight)+	an ^{-1}\left(\frac{y}{2}ight)+	an ^{-1}\left(\frac{z}{2}ight)=\frac{\pi}{2}$.Let $A = 	an ^{-1}\left(\frac{x}{2}ight)$,$B = 	an ^{-1}\left(\frac{y}{2}ight)$,and $C = 	an ^{-1}\left(\frac{z}{2}ight)$.Then $A+B+C = \frac{\pi}{2}$,which implies $A+B = \frac{\pi}{2}-C$.Taking $	an$ on both sides,we get $	an(A+B) = 	an\left(\frac{\pi}{2}-Cight) = \cot(C) = \frac{1}{	an(C)}$.Using the formula $	an(A+B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$,we have $\frac{\frac{x}{2} + \frac{y}{2}}{1 - \frac{x}{2} \cdot \frac{y}{2}} = \frac{1}{z/2}$.This simplifies to $\frac{(x+y)/2}{(4-xy)/4} = \frac{2}{z}$,which is $\frac{2(x+y)}{4-xy} = \frac{2}{z}$.Dividing by $2$,we get $\frac{x+y}{4-xy} = \frac{1}{z}$.Cross-multiplying gives $z(x+y) = 4-xy$,which means $zx + zy = 4 - xy$.Rearranging the terms,we get $xy + yz + zx = 4$.

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

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