Question:
In ∆ABC, prove the following:
$\frac{c-b \cos A}{b-c \cos A}=\frac{\cos B}{\cos C}$
Solution:
Let ABC be any triangle.
$\mathrm{LHS}=\frac{c-b \cos A}{b-c \cos A}$
$=\frac{a \cos B+b \cos A-b \cos A}{a \cos C+c \cos A-c \cos A} \quad$ [Using projection formula :
$c=a \cos B+b \cos A, b=a \cos C+c \cos A]$
$=\frac{a \cos B}{a \cos C}$
$=\frac{\cos B}{\cos C}=\mathrm{RHS}$
Hence proved.