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

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$x + z$

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

$x z$

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

answer is D.

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Detailed Solution

$T a n^{- 1} x, T a n^{- 1} y, T a n^{- 1} z a r e i n A . P .$

$\begin{array}{l} \Rightarrow 2 T a n^{- 1} y = T a n^{- 1} x + T a n^{- 1} z \\ \Rightarrow T a n^{- 1} (\frac{2 y}{1 - y^{2}}) = T a n^{- 1} (\frac{x + z}{1 - x z}) \\ \Rightarrow \frac{2 y}{1 - y^{2}} = \frac{x + z}{1 - x z} \end{array}$

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