Question

Show 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

$\frac{1+\cos (A-B)\cos C}{1+\cos (A-C)\cos B}$
$\frac{1-\cos (A-B)\cos (A+B)}{1-\cos (A-C)\cos (A+C)}$
$[\because A+B+C=\pi sothat\cos C=-\cos (A+B)]$
$=\frac{1-({\cos }^{2}A-{\sin }^{2}B)}{1-({\cos }^{2}A-{\sin }^{2}C)}$
$=\frac{{\sin }^{2}A+{\sin }^{2}B}{{\sin }^{2}A+{\sin }^{2}C}=\frac{a^2+b^2}{a^2+c^2}$.

LHS=RHS

Hence proof