In a $\Delta ABC,b\cos (C+\theta )+c\cos (B-\theta )=$

The correct option is B.$a\cos \theta$

In $△ABC,A+B+C=\pi$...(i) and

$a=2R\sin A,b=2R\sin B,c=2R\sin C$...(ii)

Now,

$b\cos (C+\theta )+c\cos (B-\theta )=R\left[2\sin B\cos \right(C+\theta )+2\sin C\cos (B-\theta \left)\right]$ [From (ii)]

$=R\left[\sin \right(B+C+\theta )+\sin (B-C-\theta )+\sin (B+C-\theta )+\sin (C-B+\theta \left)\right]$ [Using $2\sin x\sin y=\sin (x+y)+\sin (x-y)$]

$=R\left[\sin \right(B+C+\theta )+\sin (B+C-\theta )+\sin (B-C-\theta )+\sin (C-B+\theta \left)\right]$

$=R\left[\sin \right(\pi -(A-\theta ))+\sin (\pi -(A+\theta ))+\sin (B-C-\theta )-\sin (B-C-\theta \left)\right]$ [From(i)]

$=R\left[\sin \right(A-\theta )+\sin (A+\theta \left)\right]$

$=R\left[2\sin A\cos \theta \right]$

$=\left(2R\sin A\right)\cos \theta$

$=a\cos \theta$

Thus, $b\cos (C+\theta )+c\cos (B-\theta )=a\cos \theta$