If $\int \frac{\sin (x-a)}{\cos (x-b)}dx=A{\log }_e\left|\sec (x-b)\right|+Bx+C$, then B =

$\int \frac{\sin (x-a)}{\cos (x-b)}dx$

$=\int \frac{\sin (x-b-a+b)}{\cos (x-b)}dx$

$=\int \frac{\sin ((x-b)-(a-b))}{\cos (x-b)}dx$

$=\int \frac{\sin (x-b)\cos (a-b)-\cos (x-b)\sin (a-b)}{\cos (x-b)}dx$

$=\cos (a-b)\int \tan (x-b)-\sin (a-b)\int 1dx$

$=\cos (a-b)\log \left|\sec (x-b)\right|-\sin (a-b)x+C$

$B=-\sin (a-b)=\sin (b-a)$