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b-2-c-2-tan-A-c-2-a-2-tan-B-a-2-b-2-tan-C-0-

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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
b2−c2tanA+c2−a2tanB+a2−b2tanC=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 #
BythelawofSinesLetsinAa=sinBb=sinCc=ksinA=ka,sinB=kb,sinC=kcBythelawofCosinescosA=b2+c2−a22bctanA=sinAcosA=2kabcb2+c2−a2b2−c2tanA=(b2−c2)(b2+c2−a2)2kabc=b4−c4−a2b2+c2a22kabcSimilarly,c2−a2tanB=c4−a4−b2c2+a2b22kabcanda2−b2tanC=a4−b4−c2a2+b2c22kabcYou can't use 'macro parameter character #' in math mode

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