If x, y, z are in A.P. and tan^{-1}x, tan^{-1 | vedclass

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(A) Given that $x, y, z$ are in $A.P.$,we have $2y = x + z$. Since $	an^{-1}x, 	an^{-1}y, 	an^{-1}z$ are in $A.P.$,we have $2	an^{-1}y = 	an^{-1}

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$x = y = z$

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(A) Given that $x, y, z$ are in $A.P.$,we have $2y = x + z$. Since $	an^{-1}x, 	an^{-1}y, 	an^{-1}z$ are in $A.P.$,we have $2	an^{-1}y = 	an^{-1}x + 	an^{-1}z$. Using the formula $	an^{-1}A + 	an^{-1}B = 	an^{-1}\left(\frac{A+B}{1-AB}ight)$,we get $	an^{-1}\left(\frac{2y}{1-y^2}ight) = 	an^{-1}\left(\frac{x+z}{1-xz}ight)$. This implies $\frac{2y}{1-y^2} = \frac{x+z}{1-xz}$. Substituting $x+z = 2y$,we get $\frac{2y}{1-y^2} = \frac{2y}{1-xz}$. This equation holds if $2y = 0$ or $1-y^2 = 1-xz$. If $2y = 0$,then $y=0$,which implies $x+z=0$ or $x=-z$. If $1-y^2 = 1-xz$,then $y^2 = xz$. Since $x, y, z$ are in $A.P.$ ($y^2 = xz$ implies $G.P.$),the only way for numbers to be in both $A.P.$ and $G.P.$ is if $x = y = z$.

If $x, y, z$ are in $A.P.$ and $\tan^{-1}x, \tan^{-1}y, \tan^{-1}z$ are also in $A.P.$,then

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