SOLUTION: prove that for any triangle ABC, where a,b, and c are respectively the sides opposite angles A,B, and C, {{{(a*cos(A) - b*cos(B)) / (b*cos(A) - a*cos(B))}}}{{{""=""}}}{{{cos(C)}}

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Question 988624: prove that for any triangle ABC, where a,b, and c are respectively
the sides opposite angles A,B, and C,
Answer by Edwin McCravy(20060)   (Show Source): You can put this solution on YOUR website!


The law of cosines: 

, , 

when solved for the cosines, give:

, , 

So the identity becomes to show that 



On the left side multiply numerator and denominator by LCD = 2abc





Cancel terms that add to zero and collect like terms:



Rearrange the terms of the numerator to factor by grouping, 
factor 2ab out of the denominator:



Factor first two terms in numerator as the difference of squares.
Factor -c² out of last two terms in numerator:

 

Factor (b²-a²) out of numerator:





Cancel (b²-a²)'s



That is the right side of the identity.

Edwin
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