If $k(\cos A + \tan B \sin A) = \tan B$, then $k= \frac{ \sin B } { \cos(A-B) }$

Question

Given $$k(\cos A + \tan B \sin A) = \tan B$$

Show that $$k= \frac{ \sin B } { \cos(A-B) }$$

Answer 1

$$k=\frac{\tan B}{\cos A+\tan B\sin A}=\frac{\sin B}{\cos A\cos B+\sin B\sin A}=\frac{\sin B}{\cos (A-B)}.$$