Q.Prove that $\dfrac{\sin(B-C)}{\cos B\cos C}+\dfrac{\sin(C-A)}{\cos C\cos A}+\dfrac{\sin(A-B)}{\cos A\cos B}=0$

Use $\sin(X-Y)=\sin X\cos Y-\cos X\sin Y$; each fraction becomes a difference of tangents, and the three differences telescope to $0$.

To prove: $\dfrac{\sin(B-C)}{\cos B\cos C}+\dfrac{\sin(C-A)}{\cos C\cos A}+\dfrac{\sin(A-B)}{\cos A\cos B}=0.$

Step 1 — simplify the first term.

$\frac{\sin(B-C)}{\cos B\cos C}=\frac{\sin B\cos C-\cos B\sin C}{\cos B\cos C}=\frac{\sin B}{\cos B}-\frac{\sin C}{\cos C}=\tan B-\tan C.$

Step 2 — simplify the other two terms similarly. …