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

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 )))

$${if}\:{log}\left({a}+{b}+{c}\right)={loga}+{logb}+{logc} \\ $$$${prove} \\ $$$${log}\left(\frac{\mathrm{2}{a}}{\mathrm{1}−{a}^{\mathrm{2}} }+\frac{\mathrm{2}{b}}{\mathrm{1}−{b}^{\mathrm{2}} }+\frac{\mathrm{2}{c}}{\mathrm{1}−{c}^{\mathrm{2}} }\right)={log}\left(\frac{\mathrm{2}{a}}{\mathrm{1}−{a}^{\mathrm{2}} }\right)+{log}\left(\frac{\mathrm{2}{b}}{\mathrm{1}−{b}^{\mathrm{2}} }\right)+{log}\left(\frac{\mathrm{2}{c}}{\mathrm{1}−{c}^{\mathrm{2}} }\right) \\ $$

Answered by Kunal12588 last updated on 20/Apr/19

log(a+b+c)=log a + log b + log c ⇒log(a+b+c)=log(abc) ⇒a+b+c=abc let a=tan α=t_1 , b=tan β=t_2 , c=tan γ=t_3 ∴ t_1 +t_2 +t_3 =t_1 t_2 t_3 ⇒t_1 +t_2 =−t_3 (1−t_1 t_2 ) ⇒−t_3 =((t_1 +t_2 )/(1−t_1 t_2 )) ⇒tan(π−γ)=tan(α+β) ⇒α+β+γ=π ⇒2α+2β+2γ=2π ⇒tan(2α+2β)=tan(2π−2γ) ⇒((tan 2α + tan 2β)/(1−tan 2α tan2β))=−tan 2γ ⇒tan 2α + tan 2β + tan 2γ = tan 2α tan2β tan 2γ ⇒log(((2t_1 )/(1−t_1 ^2 ))+((2t_2 )/(1−t_2 ^2 ))+((2t_3 )/(1−t_3 ^2 )))=log(((2t_1 )/(1−t_1 ^2 ))×((2t_2 )/(1−t_2 ^2 ))×((2t_3 )/(1−t_3 ^2 ))) ⇒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 )))

$${log}\left({a}+{b}+{c}\right)={log}\:{a}\:+\:{log}\:{b}\:+\:{log}\:{c} \\ $$$$\Rightarrow{log}\left({a}+{b}+{c}\right)={log}\left({abc}\right) \\ $$$$\Rightarrow{a}+{b}+{c}={abc} \\ $$$${let}\:\:{a}={tan}\:\alpha={t}_{\mathrm{1}} ,\:\:{b}={tan}\:\beta={t}_{\mathrm{2}} ,\:\:{c}={tan}\:\gamma={t}_{\mathrm{3}} \\ $$$$\therefore\:\:{t}_{\mathrm{1}} +{t}_{\mathrm{2}} +{t}_{\mathrm{3}} ={t}_{\mathrm{1}} {t}_{\mathrm{2}} {t}_{\mathrm{3}} \\ $$$$\Rightarrow{t}_{\mathrm{1}} +{t}_{\mathrm{2}} =−{t}_{\mathrm{3}} \left(\mathrm{1}−{t}_{\mathrm{1}} {t}_{\mathrm{2}} \right) \\ $$$$\Rightarrow−{t}_{\mathrm{3}} =\frac{{t}_{\mathrm{1}} +{t}_{\mathrm{2}} }{\mathrm{1}−{t}_{\mathrm{1}} {t}_{\mathrm{2}} } \\ $$$$\Rightarrow{tan}\left(\pi−\gamma\right)={tan}\left(\alpha+\beta\right) \\ $$$$\Rightarrow\alpha+\beta+\gamma=\pi \\ $$$$\Rightarrow\mathrm{2}\alpha+\mathrm{2}\beta+\mathrm{2}\gamma=\mathrm{2}\pi \\ $$$$\Rightarrow{tan}\left(\mathrm{2}\alpha+\mathrm{2}\beta\right)={tan}\left(\mathrm{2}\pi−\mathrm{2}\gamma\right) \\ $$$$\Rightarrow\frac{{tan}\:\mathrm{2}\alpha\:+\:{tan}\:\mathrm{2}\beta}{\mathrm{1}−{tan}\:\mathrm{2}\alpha\:{tan}\mathrm{2}\beta}=−{tan}\:\mathrm{2}\gamma \\ $$$$\Rightarrow{tan}\:\mathrm{2}\alpha\:+\:{tan}\:\mathrm{2}\beta\:+\:{tan}\:\mathrm{2}\gamma\:=\:{tan}\:\mathrm{2}\alpha\:{tan}\mathrm{2}\beta\:\:{tan}\:\mathrm{2}\gamma \\ $$$$\Rightarrow{log}\left(\frac{\mathrm{2}{t}_{\mathrm{1}} }{\mathrm{1}−{t}_{\mathrm{1}} ^{\mathrm{2}} }+\frac{\mathrm{2}{t}_{\mathrm{2}} }{\mathrm{1}−{t}_{\mathrm{2}} ^{\mathrm{2}} }+\frac{\mathrm{2}{t}_{\mathrm{3}} }{\mathrm{1}−{t}_{\mathrm{3}} ^{\mathrm{2}} }\right)={log}\left(\frac{\mathrm{2}{t}_{\mathrm{1}} }{\mathrm{1}−{t}_{\mathrm{1}} ^{\mathrm{2}} }×\frac{\mathrm{2}{t}_{\mathrm{2}} }{\mathrm{1}−{t}_{\mathrm{2}} ^{\mathrm{2}} }×\frac{\mathrm{2}{t}_{\mathrm{3}} }{\mathrm{1}−{t}_{\mathrm{3}} ^{\mathrm{2}} }\right) \\ $$$$\Rightarrow{log}\left(\frac{\mathrm{2}{a}}{\mathrm{1}−{a}^{\mathrm{2}} }+\frac{\mathrm{2}{b}}{\mathrm{1}−{b}^{\mathrm{2}} }+\frac{\mathrm{2}{c}}{\mathrm{1}−{c}^{\mathrm{2}} }\right)={log}\left(\frac{\mathrm{2}{a}}{\mathrm{1}−{a}^{\mathrm{2}} }\right)+{log}\left(\frac{\mathrm{2}{b}}{\mathrm{1}−{b}^{\mathrm{2}} }\right)+{log}\left(\frac{\mathrm{2}{c}}{\mathrm{1}−{c}^{\mathrm{2}} }\right) \\ $$

Commented by tanmay last updated on 20/Apr/19

bah! very good excellent...

$${bah}!\:{very}\:{good}\:{excellent}... \\ $$

Commented by Kunal12588 last updated on 20/Apr/19

Thank you sir ��

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