Aplicam teorema sinusurilor:
[tex] \frac{a}{sinA} = \frac{b}{sinB}= \frac{c}{sinC}=2R\\
a=2RsinA\\
b=2RsinB\\
\frac{a^2-b^2}{a^2+b^2}= \frac{4R^2(sin^2A-sin^2B)}{4R^2(sin^2A+sin^2B)}= \\=\frac{sin^2A-sin^2B}{sin^2A+sin^2B}\\
cos2A=cos^2A-sin^2A=1-2sin^2A=\ \textgreater \ sin^2A= \frac{1-cos2A}{2}\\
\frac{sin^2A-sin^2B}{sin^2A+sin^2B}= \frac{cos2B-cos2A}{2-cos2A-2cos2B}=\\
= \frac{-2sin(A+B)sin(B-A)}{2-2cos(A+B)cos(A-B)} =\\
= \frac{2sin(A+B)sin(A-B)}{2-2cos(A+B)cos(A-B)} =\\
= \frac{sinCsin(A-B)}{1+cosCcos(A-B)} \\
Obs:A+B+C=180^0\\
[/tex]
