In a triangle $A B C$, with usual notations, $3 \mathrm{~b}=\mathrm{a}+\mathrm{c}$, then $\cot \frac{\mathrm{A}}{2} \cdot \cot \frac{\mathrm{C}}{2}=$

In a triangle ABC , with usual notations if $\frac{2 \cos \mathrm{~A}}{\mathrm{a}}+\frac{\cos \mathrm{B}}{\mathrm{b}}+\frac{2 \cos \mathrm{C}}{\mathrm{c}}=\frac{\mathrm{a}}{\mathrm{bc}}+\frac{\mathrm{b}}{\mathrm{ca}}$ then $\angle \mathrm{A}=$

In a triangle ABC with usual notations, if $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are in arithmetic progression, then, $\tan \frac{\mathrm{A}}{2} \cdot \tan \frac{\mathrm{C}}{2}=$

With usual notations, in a triangle $A B C$, if $\theta$ is any real number, then $a \cos (B-\theta)+b \cos (A+\theta)$ is