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Question Number 21266 by oyshi last updated on 18/Sep/17

((b^2 −c^2 )/(tan A))+((c^2 −a^2 )/(tan B))+((a^2 −b^2 )/(tan C))=0

$$\frac{{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{tan}\:{A}}+\frac{{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{tan}\:{B}}+\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{tan}\:{C}}=\mathrm{0} \\ $$

Answered by myintkhaing last updated on 18/Sep/17

By the law of Sines Let ((sinA)/a) = ((sinB)/b) = ((sinC)/c) = k sinA= ka, sinB=kb, sinC=kc By the law of Cosines cosA=((b^2 +c^2 −a^2 )/(2bc)) tanA=((sinA)/(cosA)) = ((2kabc)/(b^2 +c^2 −a^2 )) ((b^2 −c^2 )/(tanA))=(((b^2 −c^2 )(b^2 +c^2 −a^2 ))/(2kabc))=((b^4 −c^4 −a^2 b^2 +c^2 a^2 )/(2kabc)) Similarly, ((c^2 −a^2 )/(tanB)) =((c^4 −a^4 −b^2 c^2 +a^2 b^2 )/(2kabc)) and ((a^2 −b^2 )/(tanC)) = ((a^4 −b^4 −c^2 a^2 +b^2 c^2 )/(2kabc)) ((b^2 −c^2 )/(tanA))+((c^2 −a^2 )/(tanB))+((a^2 −b^2 )/(tanC)) = 0 #

$$\mathrm{By}\:\mathrm{the}\:\mathrm{law}\:\mathrm{of}\:\mathrm{Sines} \\ $$$$\mathrm{Let}\:\frac{\mathrm{sinA}}{\mathrm{a}}\:=\:\frac{\mathrm{sinB}}{\mathrm{b}}\:=\:\frac{\mathrm{sinC}}{\mathrm{c}}\:=\:\mathrm{k} \\ $$$$\mathrm{sinA}=\:\mathrm{ka},\:\mathrm{sinB}=\mathrm{kb},\:\mathrm{sinC}=\mathrm{kc} \\ $$$$\mathrm{By}\:\mathrm{the}\:\mathrm{law}\:\mathrm{of}\:\mathrm{Cosines} \\ $$$$\mathrm{cosA}=\frac{\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} }{\mathrm{2bc}} \\ $$$$\mathrm{tanA}=\frac{\mathrm{sinA}}{\mathrm{cosA}}\:=\:\frac{\mathrm{2kabc}}{\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{b}^{\mathrm{2}} −\mathrm{c}^{\mathrm{2}} }{\mathrm{tanA}}=\frac{\left(\mathrm{b}^{\mathrm{2}} −\mathrm{c}^{\mathrm{2}} \right)\left(\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} \right)}{\mathrm{2kabc}}=\frac{\mathrm{b}^{\mathrm{4}} −\mathrm{c}^{\mathrm{4}} −\mathrm{a}^{\mathrm{2}} \mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} \mathrm{a}^{\mathrm{2}} }{\mathrm{2kabc}} \\ $$$$\mathrm{Similarly}, \\ $$$$\frac{\mathrm{c}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} }{\mathrm{tanB}}\:=\frac{\mathrm{c}^{\mathrm{4}} −\mathrm{a}^{\mathrm{4}} −\mathrm{b}^{\mathrm{2}} \mathrm{c}^{\mathrm{2}} +\mathrm{a}^{\mathrm{2}} \mathrm{b}^{\mathrm{2}} }{\mathrm{2kabc}}\:\mathrm{and} \\ $$$$\frac{\mathrm{a}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}} }{\mathrm{tanC}}\:=\:\frac{\mathrm{a}^{\mathrm{4}} −\mathrm{b}^{\mathrm{4}} −\mathrm{c}^{\mathrm{2}} \mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} \mathrm{c}^{\mathrm{2}} }{\mathrm{2kabc}} \\ $$$$\frac{\mathrm{b}^{\mathrm{2}} −\mathrm{c}^{\mathrm{2}} }{\mathrm{tanA}}+\frac{\mathrm{c}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} }{\mathrm{tanB}}+\frac{\mathrm{a}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}} }{\mathrm{tanC}}\:=\:\mathrm{0}\:# \\ $$

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