If tan-1x, tan-1y,tan-1z are in A.P. then 2y1-y2=

tan-1x, tan-1y, tan-1z=→A.P 2tan-1y=tan-1x+tan-1z tan-12y1-y2=tan-1x+z1-xz 2y1-y2=x+z1-xz

If tan-1x, tan-1y,tan-1z are in A.P. then 2y1-y2=

tan-1x, tan-1y, tan-1z=→A.P 2tan-1y=tan-1x+tan-1z tan-12y1-y2=tan-1x+z1-xz 2y1-y2=x+z1-xz

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If $\tan^{- 1} x, \tan^{- 1} y, \tan^{- 1} z$ are in A.P. then $\frac{2 y}{1 - y^{2}} =$

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$\frac{x - z}{1 + xz}$

$\frac{x + z}{1 - xz}$

x + z

answer is B.

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Detailed Solution

$\tan^{- 1} x, \tan^{- 1} y, \tan^{- 1} z = \to A . P 2 \tan^{- 1} y = \tan^{- 1} x + \tan^{- 1} z \tan^{- 1} (\frac{2 y}{1 - y^{2}}) = \tan^{- 1} (\frac{x + z}{1 - xz}) \frac{2 y}{1 - y^{2}} = \frac{x + z}{1 - xz}$

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