In $$\Delta ABC$$, with usual notations $$\mathrm{\frac{b\sin B-c\sin C}{\sin(B-C)}}=$$

With usual notations, in any $$\triangle A B C$$, if $$a\cos B=b \cos A$$, then the triangle is

In $$\triangle A B C$$, with usual notations, $$2 a b \sin \frac{1}{2}(A+B-C)=$$

If in $$\Delta$$ABC, with usual notations, the angles are in A.P., then $$\mathrm{\frac{a}{c}}$$ sin 2 C + $$\mathrm{\frac{c}{a}}$$ sin 2 A =