if log(a+b+c)=loga+logb+logc prove log(((2a)/(1−a^2 ))+((2b)/(1−b^2 ))+((2c)/(1−c^2 )))=log(((2a)/(1−a^2 )))+log(((2b)/(1−b^2 )))+log(((2c)/(1−c^2 )))


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Question Number 58245 by tanmay last updated on 20/Apr/19

$i f l o g (a + b + c) = l o g a + l o g b + l o g c$ $p r o v e$ $l o g (\frac{2 a}{1 - a^{2}} + \frac{2 b}{1 - b^{2}} + \frac{2 c}{1 - c^{2}}) = l o g (\frac{2 a}{1 - a^{2}}) + l o g (\frac{2 b}{1 - b^{2}}) + l o g (\frac{2 c}{1 - c^{2}})$
	Answered by Kunal12588 last updated on 20/Apr/19

	$l o g (a + b + c) = l o g a + l o g b + l o g c$ $\Rightarrow l o g (a + b + c) = l o g (a b c)$ $\Rightarrow a + b + c = a b c$ $l e t a = t a n α = t_{1}, b = t a n β = t_{2}, c = t a n γ = t_{3}$ $∴ t_{1} + t_{2} + t_{3} = t_{1} t_{2} t_{3}$ $\Rightarrow t_{1} + t_{2} = - t_{3} (1 - t_{1} t_{2})$ $\Rightarrow - t_{3} = \frac{t_{1} + t_{2}}{1 - t_{1} t_{2}}$ $\Rightarrow t a n (π - γ) = t a n (α + β)$ $\Rightarrow α + β + γ = π$ $\Rightarrow 2 α + 2 β + 2 γ = 2 π$ $\Rightarrow t a n (2 α + 2 β) = t a n (2 π - 2 γ)$ $\Rightarrow \frac{t a n 2 α + t a n 2 β}{1 - t a n 2 α t a n 2 β} = - t a n 2 γ$ $\Rightarrow t a n 2 α + t a n 2 β + t a n 2 γ = t a n 2 α t a n 2 β t a n 2 γ$ $\Rightarrow l o g (\frac{2 t_{1}}{1 - t_{1}^{2}} + \frac{2 t_{2}}{1 - t_{2}^{2}} + \frac{2 t_{3}}{1 - t_{3}^{2}}) = l o g (\frac{2 t_{1}}{1 - t_{1}^{2}} \times \frac{2 t_{2}}{1 - t_{2}^{2}} \times \frac{2 t_{3}}{1 - t_{3}^{2}})$ $\Rightarrow l o g (\frac{2 a}{1 - a^{2}} + \frac{2 b}{1 - b^{2}} + \frac{2 c}{1 - c^{2}}) = l o g (\frac{2 a}{1 - a^{2}}) + l o g (\frac{2 b}{1 - b^{2}}) + l o g (\frac{2 c}{1 - c^{2}})$
	Commented by tanmay last updated on 20/Apr/19

	$b a h! v e r y g o o d e x c e l l e n t . . .$
	Commented by Kunal12588 last updated on 20/Apr/19
	Thank you sir ��
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