Prove that:
\[\frac{\sin \left( A - B \right)}{\cos A \cos B} + \frac{\sin \left( B - C \right)}{\cos B \cos C} + \frac{\sin \left( C - A \right)}{\cos C \cos A} = 0\]
Solution
\[\text{ LHS }= \frac{\sin\left( A - B \right)}{\cos A \cos B} + \frac{\sin\left( B - C \right)}{\cos B \cos C} + \frac{\sin\left( C - A \right)}{\cos C \cos A}\]
\[ = \frac{\sin A \cos B - \cos A \sin B}{\cos A \cos B} + \frac{\sin B \cos C - \cos B \sin C}{\cos B \cos C} + \frac{\sin C \cos A - \cos C \sin A}{\cos C \cos A}\]
\[ = \frac{\sin A \cos B}{\cos A \cos B} - \frac{\cos A \sin B}{\cos A \cos B} + \frac{\sin B \cos C}{\cos B \cos C} - \frac{\cos B \sin C}{\cos B \cos C} + \frac{\sin C \cos A}{\cos C \cos A} - \frac{\cos C \sin A}{\cos C \cos A}\]
\[ = \frac{\sin A}{\cos A} - \frac{\sin B}{\cos B} + \frac{\sin B}{\cos B} - \frac{\sin C}{\cos C} + \frac{\sin C}{\cos C} - \frac{\sin A}{\cos A}\]
\[ = \tan A - \tan B + \tan B - \tan C + \tan C - \tan A\]
\[ = 0\]
= RHS
Hence proved.
