Question

Prove that $\frac{1+\cos (A-B)\cos C}{1+\cos (A-C)\cos B}=\frac{a^2+b^2}{a^2+c^2}$

Answer 1

L.H.S.$=\frac{1+\cos (A-B)\cos C}{1+\cos (A-C)\cos B}=\frac{1-\cos (A-B)\cos (A+B)}{1-\cos (A-C)\cos (A+C)}$
$=\frac{1-{\cos }^{2}A+{\sin }^{2}B}{1-{\cos }^{2}A+{\sin }^{2}C}=\frac{{\sin }^{2}A+{\sin }^{2}B}{{\sin }^{2}A+{\sin }^{2}C}$ = $\frac{a^2+b^2}{a^2+c^2}$ [using $a=2R\sin A$,etc.]