Question

If in triangle $ABC,$ sides opposite to angles $A,B,C$ are $a,b$ and $c$ respectively, then the value of $\frac{1+\cos (A-B)\cos C}{1+\cos (A-C)\cos B}=$

• A. $\frac{a-b}{b-c}$
• B. $\frac{a+b}{a-b}$
• C. $\frac{a^2-b^2}{a^2-c^2}$
• D. $\frac{a^2+b^2}{a^2+c^2}$

Answer 1

The correct option is D $\frac{a^2+b^2}{a^2+c^2}$

$\Rightarrow \frac{1+\cos (A-B)\cos C}{1+\cos (A-C)\cos B}=\frac{1+\frac{\cos (A-B+C)+\cos (A-B-C)}{2}}{1+\frac{\cos (A-C+B)+\cos (A-C-B)}{2}}=\frac{2+\cos (\pi -2B)+\cos (A-(\pi -A\left)\right)}{2+\cos (\pi -2C)+\cos (A-(\pi -A\left)\right)}=\frac{2-\cos 2B-\cos 2A}{2-\cos 2C-\cos 2A}$
$=\frac{2{\sin }^{2}B+2{\sin }^{2}A}{2{\sin }^{2}A+2{\sin }^{2}C}$
Using, sine rule
$=\frac{a^2+b^2}{a^2+c^2}$