In ∆ABC, prove that if θ be any angle,

Suppose $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$.         ....(1)

Consider the RHS of the equation b cosθ = c cos (A − θ) + a cos (C + θ).

RHS $=c \cos (A-\theta)+a \cos (C+\theta)$

$=k \sin C \cos (A-\theta)+k \sin A \cos (C+\theta) \quad$ (from (1))

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

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

$=\frac{k}{2}(\sin (\pi-\mathrm{B}-\theta)+\sin (C+\theta-A)+\sin (\pi-\mathrm{B}+\theta)-\sin (C+\theta-A))$   $(\because \mathrm{A}+\mathrm{B}+\mathrm{C}=\pi)$      

$=\frac{k}{2}(\sin (B+\theta)+\sin (B-\theta))$

$=\frac{k}{2}(\sin B \cos \theta+\sin \theta \cos B+\sin B \cos \theta-\sin \theta \cos B)$

$=\frac{k}{2}(2 \sin B \cos \theta)$

$=k \sin B \cos \theta$

$=b \cos \theta=\mathrm{LHS}$

Hence proved.