In triangle ABC, prove the following:

Let $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k \quad \ldots$ (1)

Consider the LHS of the equation $a \cos A+b \cos B+c \cos C$.

$a \cos A+b \cos B+c \cos C=k(\sin A \cos A+\sin B \cos B+\sin C \cos C)$

$=\frac{k}{2}(2 \sin A \cos A+2 \sin A \cos A+2 \sin C \cos C)$

$=\frac{k}{2}(\sin 2 A+\sin 2 B+\sin 2 C)$

$=\frac{k}{2}[2 \sin (A+B) \cos (A-B)+2 \sin C \cos C]$

$=\frac{k}{2}[2 \sin (\pi-C) \cos (A-B)+2 \sin C \cos C]$

$=\frac{k}{2}[2 \sin C \cos (A-B)+2 \sin C \cos C]$

$=\frac{2 k \sin C}{2}[\cos (A-B)+\cos C]$

$=k \sin C[\cos (A-B)+\cos \{\pi-(\mathrm{A}+\mathrm{B})\}]$

$=k \sin C[\cos (A-B)-\cos (\mathrm{A}+\mathrm{B})]$

$=k \sin C[2 \sin A \sin B]$

$=2 k \sin A \sin B \sin C \quad \ldots(1)$

Now,

on putting $k \sin C=C$ in equation (1), we get:

$2 c \sin A \sin B$

and on putting $k \sin B=b$ in equation (1), we get:

$2 b \sin A \sin C$

So, from (1), we have

$a \cos A+b \cos B+c \cos C=2 b \sin A \sin C=2 c \sin A \sin B$

Hence proved.