a (cos B cos C+cos A)=b

Solution:

Suppose $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$

Consider:

$a(\cos B \cos C+\cos A)$

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

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

$=k\left[\frac{1}{2} \cos C\{\sin (A+B)+\sin (A-B)\}+\sin A \cos A\right]$

$=k\left[\frac{1}{2}\{\sin (A+B) \cos C+\sin (A-B) \cos C\}+\sin A \cos A\right]$

$=k\left[\frac{1}{2}\left\{\frac{1}{2}[\sin (A+B+C)+\sin (A+B-C)+\sin (A-B+C)+\sin (A-B-C)]\right\}+\sin A \cos A\right]$

$=k\left[\frac{1}{4}\{\sin \pi+\sin (\pi-2 \mathrm{C})+\sin (\pi-2 \mathrm{~B})-\sin (\pi-2 \mathrm{~A})\}+\frac{\sin 2 \mathrm{~A}}{2}\right] \quad(\because \mathrm{A}+\mathrm{B}+\mathrm{C}=\pi)$

$=\frac{k}{4}(\sin 2 C+\sin 2 B+\sin 2 A) \ldots$ (1)

and,

$b(\cos A \cos C+\cos B)$

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

$=k\left[\frac{1}{2} \cos A\{\sin (B+C)+\sin (B-C)\}+\frac{\sin 2 B}{2}\right]$

$=k\left(\frac{1}{2}(\sin (B+C) \cos A+\sin (B-C) \cos A)+\frac{\sin 2 B}{2}\right)$

$=k\left(\frac{1}{4}(\sin (B+C+A)+\sin (B+C-A)+\sin (B-C+A)+\sin (B-C-A))+\frac{\sin 2 B}{2}\right)$

$=\frac{k}{4}\left(\sin \pi+\sin (\pi-2 \mathrm{~A})+\sin (\pi-2 \mathrm{C})-\sin (\pi-2 \mathrm{~B})+\frac{\sin 2 \mathrm{~B}}{2}\right) \quad(\because \mathrm{A}+\mathrm{B}+\mathrm{C}=\pi)$

$=\frac{k}{4}(\sin 2 A+\sin 2 C+\sin 2 B) \ldots(2)$

Similarly,

$c(\cos A \cos B+\cos C)=\frac{k}{4}(\sin 2 A+\sin 2 B+\sin 2 C) \quad \ldots(3)$

From (1), (2) and (3), we get:

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

Hence proved.