If sin(x-y) = sin x cos y - cos x sin y, then the value of $$\frac{\sin(A - B)}{\cos A \cos B} + \frac{\sin(B - C)}{\cos B \cos C}$$​is:

Correct option is B

Given:
sin(x-y) = sin x cos y - cos x sin y
Formula Used:
​$$\frac{\sin(A - B)}{\cos A \cos B} + \frac{\sin(B - C)}{\cos B \cos C} \\\ \\= \frac{\sin A \cos B - \cos A \sin B}{\cos A \cos B} + \frac{\sin B \cos C - \cos B \sin C}{\cos B \cos C} \\\ \\= \frac{\sin A \cos B}{\cos A \cos B} - \frac{\cos A \sin B}{\cos A \cos B} + \frac{\sin B \cos C}{\cos B \cos C} - \frac{\cos B \sin C}{\cos B \cos C} \\\ \\= \frac{\sin A}{\cos A} - \frac{\sin B}{\cos B} + \frac{\sin B}{\cos B} - \frac{\sin C}{\cos C} \\\ \\= \frac{\sin A}{\cos A} - \frac{\sin C}{\cos C} \\\ \\= \frac{\sin A \cos C - \cos A \sin C}{\cos A \cos C} \\\ \\= \frac{\sin(A - C)}{\cos A \cos C}$$​​