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

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(A) Given that $x, y, z$ are in $A.P.$,we have $2y = x + z$. Also,$	an^{-1} x, 	an^{-1} y, 	an^{-1} z$ are in $A.P.$,so $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 $2y = x + z$. Also,$	an^{-1} x, 	an^{-1} y, 	an^{-1} z$ are in $A.P.$,so $2 	an^{-1} y = 	an^{-1} x + 	an^{-1} z$. 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 $2y = 0$ (which implies $x = -z$,but then $y=0$ and $x, y, z$ would be $x, 0, -x$,which satisfies the condition) or if $1-y^2 = 1-xz$,which means $y^2 = xz$. If $x, y, z$ are in $A.P.$ and $G.P.$,then $x = y = z$.

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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