Question

In a triangle $ABC;\frac{a(a+c-b)}{b(b+c-a)}$ is equal to

• A. $\frac{1-\cos A}{1-\cos B}$
• B. $\frac{1+\cos A}{1+\cos B}$
• C. $\frac{{\cos }^2\left(\frac{A}{2}\right)}{{\sin }^2\left(\frac{B}{2}\right)}$
• D. $\frac{{\sin }^{2}A}{{\sin }^{2}B}$

Answer 1

The correct option is A $\frac{1-\cos A}{1-\cos B}$

$\frac{a(a+c-b)}{b(b+c-a)}$
Multiplying and dividing by $(a+b+c)$, we get
$=\frac{a\left[\right(a+c{)}^2-b^2]}{b\left[\right(b+c{)}^2-a^2]}$
$=\frac{a[a^2+c^2-b^2+2ac]}{b[b^2+c^2-a^2+2bc]}$
$=\frac{a\left[2ac\right(\cos B+1\left)\right]}{b\left[2bc\right(\cos A+1\left)\right]}$
$=\frac{a^2(1+\cos B)}{b^2(1+\cos A)}$
$=\frac{a^2{\sin }^{2}B.(1-\cos A)}{b^2{\sin }^{2}A(1-\cos B)}$
Now by sine rule
$\frac{a}{\sin A}=\frac{b}{\sin B}=k$
Hence
$\frac{a^2}{{\sin }^{2}A}.\frac{{\sin }^{2}B}{b^2}.\frac{1-\cos A}{1-\cos B}$
$=\frac{1-\cos A}{1-\cos B}$.