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

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(D) Given that $x, y, z$ are in $G$.$P$.,we have $y^2 = xz$ $(i)$.Also,$	an^{-1} x, 	an^{-1} y, 	an^{-1} z$ are in $A$.$P$.,so $2 	an^{-1} y = 	a

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$x = y = z$

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(D) Given that $x, y, z$ are in $G$.$P$.,we have $y^2 = xz$ $(i)$.Also,$	an^{-1} x, 	an^{-1} y, 	an^{-1} z$ are in $A$.$P$.,so $2 	an^{-1} y = 	an^{-1} x + 	an^{-1} z$.Applying 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 $\frac{2y}{1-y^2} = \frac{x+z}{1-xz}$.Since $y^2 = xz$,the denominators are equal,so $2y = x+z$ (ii).From $(i)$ and (ii),$x, y, z$ are in both $A$.$P$. and $G$.$P$.,which implies $x = y = z$.

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

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