If x, y, z are in arithmetic progression and | vedclass

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Question

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

Vedclass · Accepted Answer

$x = y = z$

Vedclass · Answer

(A) Given that $x, y, z$ are in arithmetic progression,we have $2y = x + z$. Also,$	an^{-1}x, 	an^{-1}y, 	an^{-1}z$ are in arithmetic progression,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 $\frac{2y}{1-y^2} = \frac{x+z}{1-xz}$. Since $x+z = 2y$,we have $\frac{2y}{1-y^2} = \frac{2y}{1-xz}$. This equation holds if $2y = 0$ or $1-y^2 = 1-xz$. If $2y = 0$,then $y = 0$,which implies $x = -z$. If $1-y^2 = 1-xz$,then $y^2 = xz$. Since $x, y, z$ are in arithmetic progression,$y^2 = xz$ implies $x = y = z$. Thus,the condition $x = y = z$ satisfies the given requirements.

If $x, y, z$ are in arithmetic progression and $\tan^{-1}x, \tan^{-1}y, \tan^{-1}z$ are also in arithmetic progression,then:

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