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

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$x=y=z$

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(A) Given that $x, y, z$ are in $A.P.$$	herefore 2y = x + z$ ... $(i)$Also,$	an ^{-1} x, 	an ^{-1} y, 	an ^{-1} z$ are in $A.P.$$	herefore 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)$:$	an ^{-1} \left( \frac{2y}{1-y^2} ight) = 	an ^{-1} \left( \frac{x+z}{1-xz} ight)$$\frac{2y}{1-y^2} = \frac{x+z}{1-xz}$Substituting $x+z = 2y$ from $(i)$:$\frac{2y}{1-y^2} = \frac{2y}{1-xz}$This implies either $2y = 0$ (which leads to $x=y=z=0$) or $1-y^2 = 1-xz$,which gives $y^2 = xz$.Since $x, y, z$ are in both $A.P.$ and $G.P.$,we must have $x=y=z$.

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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