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 $2y = x + z$ ....$(1)$.Also,$	an^{-1} x, 	an^{-1} y, 	an^{-1} z$ are in $A.P.$,so $2 	an^{-1} y = \

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$x = y = z$

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(A) Given that $x, y, z$ are in $A.P.$,we have $2y = x + z$ ....$(1)$.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 $2 	an^{-1} y = 	an^{-1} \left( \frac{x+z}{1-xz} ight)$.Taking $	an$ on both sides: $\frac{2y}{1-y^2} = \frac{x+z}{1-xz}$.Substituting $x+z = 2y$ from $(1)$,we get $\frac{2y}{1-y^2} = \frac{2y}{1-xz}$.This implies $2y = 0$ or $\frac{1}{1-y^2} = \frac{1}{1-xz}$.If $2y = 0$,then $y = 0$,which implies $x+z = 0$,so $z = -x$. This satisfies the condition $x, y, z$ in $A.P.$ $(x, 0, -x)$.If $1-y^2 = 1-xz$,then $y^2 = xz$. Since $x, y, z$ are in $A.P.$ and $G.P.$,they must be equal,i.e.,$x = y = z$.Comparing the options,$x = y = z$ is the standard result.

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

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