$If x + y - z + x y z = 0, then$ $\frac{2 x}{1 - x^{2}} + \frac{2 y}{1 - y^{2}} - \frac{2 z}{1 - z^{2}} is equal$ to

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$\frac{x y z}{[(1 - x^{2}) (1 - y^{2}) (1 - z^{2})]}$

$\frac{- 8 x y z}{[(1 - x^{2}) (1 - y^{2}) (1 - z^{2})]}$

$\frac{- x y z}{[(1 - x^{2}) (1 - y^{2}) (1 - z^{2})]}$

$\frac{8 x y z}{[(1 - x^{2}) (1 - y^{2}) (1 - z^{2})]}$

answer is C.

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Detailed Solution

$\begin{array}{l} L e t x = \tan A, y = \tan B, - z = \tan C . Then (x + y - z) = - x y z \\ \Rightarrow \tan A + \tan B + \tan C = \tan A \tan B \tan C \\ \Rightarrow A + B + C = π \Rightarrow 2 A + 2 B = 2 π - 2 C \\ \Rightarrow \tan (2 A + 2 B) = \tan (2 π - 2 C) = - \tan 2 C \\ \Rightarrow \tan 2 A + \tan 2 B + \tan 2 C = \tan 2 A \cdot \tan 2 B \cdot \tan 2 C \\ \Rightarrow \frac{2 \tan A}{1 - \tan^{2} A} + \frac{2 \tan B}{1 - \tan^{2} B} + \frac{2 \tan C}{1 - \tan^{2} C} = \frac{2 \tan A}{1 - \tan^{2} A} \cdot \frac{2 \tan B}{1 - \tan^{2} B} \cdot \frac{2 \tan C}{1 - \tan^{2} C} \\ Put the value of \tan A, \tan B, \tan C, we get \\ \Rightarrow \frac{2 x}{1 - x^{2}} + \frac{2 y}{1 - y^{2}} - \frac{2 z}{1 - z^{2}} = - \frac{8 x y z}{(1 - x^{2}) (1 - y^{2}) (1 - z^{2})} \end{array}$