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-

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 \dots

Commented by Kunal12588 last updated on 20/Apr/19

Thank you sir ��