Question

If three angles $A,B$ and $C$ are angles of a cyclic quadrilateral, prove that $\cot B=\frac{\sin A-\sin C}{\cos C-\cos A}$

Answer 1

RHS $=\frac{2\sin \left\{(A-C)/2\right\}\cos \left\{(A+C)/2\right\}}{2\sin \left\{(C+A)/2\right\}\sin \left\{(A-C)/2\right\}}=\cot \left\{(A+C)/2\right\}=\cot B$