Question 988624
<pre>
{{{(a*cos(A) - b*cos(B)) / (b*cos(A) - a*cos(B))}}}{{{""=""}}}{{{cos(C)}}}

The law of cosines: 

{{{a^2 = b^2+c^2-2ab*cos(A)}}}, {{{b^2 = a^2+c^2-2ac*cos(B)}}}, {{{c^2=b^2+a^2-2ab*cos(C)}}}

when solved for the cosines, give:

{{{cos(A)=(b^2+c^2-a^2)/(2bc)}}}, {{{cos(B)=(a^2+c^2-b^2)/(2ac)}}}, {{{cos(C)=(b^2+a^2-c^2)/(2ab)}}}

So the identity becomes to show that 

{{{(a*((b^2+c^2-a^2)/(2bc)) - b*((a^2+c^2-b^2)/(2ac))) / (b*((b^2+c^2-a^2)/(2bc)) - a*((a^2+c^2-b^2)/(2ac)))}}}{{{""=""}}}{{{(b^2+a^2-c^2)/(2ab)}}}

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

{{{ ( a^2(b^2+c^2-a^2) - b^2(a^2+c^2-b^2) ) / (ab(b^2+c^2-a^2) - ab(a^2+c^2-b^2) )}}}

{{{ (  a^2b^2+a^2c^2-a^4 - b^2a^2-b^2c^2+b^4 ) / (ab^3+abc^2-a^3b - a^3b-abc^2+ab^3) }}}

Cancel terms that add to zero and collect like terms:

{{{ (  a^2c^2-a^4-b^2c^2+b^4 ) / (2ab^3-2a^3b) }}}

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

{{{ ( b^4-a^4-b^2c^2+a^2c^2 ) / (2ab(b^2-a^2)) }}}

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

{{{ ( (b^2-a^2)(b^2+a^2)-c^2(b^2-a^2) ) / (2ab(b^2-a^2)) }}} 

Factor (b²-a²) out of numerator:

{{{ ( (b^2-a^2)((b^2+a^2)^""-c^2) ) / (2ab(b^2-a^2)) }}}

{{{ ( (b^2-a^2)(b^2+a^2-c^2) ) / (2ab(b^2-a^2)) }}}

Cancel (b²-a²)'s

{{{ (b^2+a^2-c^2) / (2ab) }}}

That is the right side of the identity.

Edwin</pre>