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Question Number 75209 by vishalbhardwaj last updated on 08/Dec/19

	Answered by Kunal12588 last updated on 08/Dec/19

	$x + y + z = x y z$ $l e t x = t a n α, y = t a n β, z = t a n γ$ $t a n α + t a n β + t a n γ = t a n α t a n β t a n γ$ $\Rightarrow t a n γ (1 - t a n α t a n β) = - (t a n α + t a n β)$ $\Rightarrow t a n γ = - \frac{t a n α + t a n β}{1 - t a n α t a n β} = - t a n (α + β)$ $\Rightarrow t a n (α + β) = - t a n γ = t a n (π - γ)$ $\Rightarrow α + β = π - γ$ $\Rightarrow α + β + γ = π$ $\Rightarrow 2 α + 2 β = 2 π - 2 γ$ $\Rightarrow t a n (2 α + 2 β) = t a n (2 π - 2 γ)$ $\Rightarrow \frac{t a n 2 α + t a n 2 β}{1 - t a n 2 α t a n 2 β} = - t a n 2 γ$ $\Rightarrow t a n 2 α + t a n 2 β = - t a n 2 γ + t a n 2 α t a n 2 β t a n 2 γ$ $\Rightarrow t a n 2 α + t a n 2 β + t a n 2 γ = t a n 2 α t a n 2 β t a n 2 γ$ $\Rightarrow \frac{2 t a n α}{1 - {t a n}^{2} α} + \frac{2 t a n β}{1 - {t a n}^{2} β} + \frac{2 t a n γ}{1 - {t a n}^{2} γ} = \frac{2 t a n α}{1 - {t a n}^{2} α} \times \frac{2 t a n β}{1 - {t a n}^{2} β} \times \frac{2 t a n γ}{1 - {t a n}^{2} γ}$ $\Rightarrow \frac{2 x}{1 - x^{2}} + \frac{2 y}{1 - y^{2}} + \frac{2 z}{1 - z^{2}} = (\frac{2 x}{1 - x^{2}}) (\frac{2 y}{1 - y^{2}}) (\frac{2 z}{1 - z^{2}})$
	Answered by Kunal12588 last updated on 08/Dec/19

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