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

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Question

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

Vedclass · Accepted Answer

$x = y = z$

Vedclass · Answer

$2 y=x+z$         ....$(1)$

As $	an ^{-1} x, 	an ^{-1} y, 	an ^{-1} z$ in $A.P$

$\Rightarrow 2 	an ^{-1} y=	an ^{-1} \frac{x+z}{1-x z}$

$\frac{2 y}{1-y^{2}}=\frac{x+z}{1-x z}$

$\frac{x-z}{1-y^{2}}=\frac{x+z}{1-x z}$ by $( 1)$

$(x+z)\left\{\frac{1}{1-y^{2}}-\frac{1}{1-x z}ight\}=0$

$(x+z)=0$ or $1-x z=x-y^{2}$

$y^{2}=x z$

$\Rightarrow x, y, z$ in $GP.$

As $x, y, z$  $AP$  and  $ G P$

$\Rightarrow x=y=z$

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

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