Prove that: sin(A−B)/cosAcosB + sin(B−C)/cosBcosC + sin(C−A)/cosCcosA = 0

Question

Prove that:

\[ \frac{\sin(A-B)}{\cos A\cos B} + \frac{\sin(B-C)}{\cos B\cos C} + \frac{\sin(C-A)}{\cos C\cos A} =0 \]

Proof

L.H.S.

\[ = \frac{\sin(A-B)}{\cos A\cos B} + \frac{\sin(B-C)}{\cos B\cos C} + \frac{\sin(C-A)}{\cos C\cos A} \]

Using the identity:

\[ \sin(X-Y) = \sin X\cos Y-\cos X\sin Y \]

we get:

\[ = \frac{\sin A\cos B-\cos A\sin B}{\cos A\cos B} \]

\[ + \frac{\sin B\cos C-\cos B\sin C}{\cos B\cos C} \]

\[ + \frac{\sin C\cos A-\cos C\sin A}{\cos C\cos A} \]

Separating the terms:

\[ = \frac{\sin A\cos B}{\cos A\cos B} – \frac{\cos A\sin B}{\cos A\cos B} \]

\[ + \frac{\sin B\cos C}{\cos B\cos C} – \frac{\cos B\sin C}{\cos B\cos C} \]

\[ + \frac{\sin C\cos A}{\cos C\cos A} – \frac{\cos C\sin A}{\cos C\cos A} \]

\[ = \tan A-\tan B+\tan B-\tan C+\tan C-\tan A \]

All terms cancel out:

\[ =0 \]

R.H.S.

\[ =0 \]

Hence,

\[ \frac{\sin(A-B)}{\cos A\cos B} + \frac{\sin(B-C)}{\cos B\cos C} + \frac{\sin(C-A)}{\cos C\cos A} =0 \]

Hence proved.