Prove that: sin(A−B)/sinA sinB + sin(B−C)/sinB sinC + sin(C−A)/sinC sinA = 0
Question
Prove that:
\[ \frac{\sin(A-B)}{\sin A\sin B} + \frac{\sin(B-C)}{\sin B\sin C} + \frac{\sin(C-A)}{\sin C\sin A} =0 \]
Proof
L.H.S.
\[ = \frac{\sin(A-B)}{\sin A\sin B} + \frac{\sin(B-C)}{\sin B\sin C} + \frac{\sin(C-A)}{\sin C\sin A} \]
Using the identity:
\[ \sin(X-Y) = \sin X\cos Y-\cos X\sin Y \]
we get:
\[ = \frac{\sin A\cos B-\cos A\sin B}{\sin A\sin B} \]
\[ + \frac{\sin B\cos C-\cos B\sin C}{\sin B\sin C} \]
\[ + \frac{\sin C\cos A-\cos C\sin A}{\sin C\sin A} \]
Separating the terms:
\[ = \frac{\sin A\cos B}{\sin A\sin B} – \frac{\cos A\sin B}{\sin A\sin B} \]
\[ + \frac{\sin B\cos C}{\sin B\sin C} – \frac{\cos B\sin C}{\sin B\sin C} \]
\[ + \frac{\sin C\cos A}{\sin C\sin A} – \frac{\cos C\sin A}{\sin C\sin A} \]
\[ = \cot B-\cot A+\cot C-\cot B+\cot A-\cot C \]
All terms cancel out:
\[ =0 \]
R.H.S.
\[ =0 \]
Hence,
\[ \frac{\sin(A-B)}{\sin A\sin B} + \frac{\sin(B-C)}{\sin B\sin C} + \frac{\sin(C-A)}{\sin C\sin A} =0 \]
Hence proved.