If $\tan \frac{B - C}{2} = x \cot \frac{A}{2},$ then $x = $

If in any $\Delta ABC$,$\cot \frac{A}{2}, \cot \frac{B}{2}, \cot \frac{C}{2}$ are in $A.P.$,then:

In a $\Delta ABC$,$\frac{\cos^2 \left( \frac{B - C}{2} \right)}{(b + c)^2} + \frac{\sin^2 \left( \frac{B - C}{2} \right)}{(b - c)^2} = $ (in $/ a^2$)

In a $\Delta ABC$,if ${c^2} + {a^2} - {b^2} = ac$,then $\angle B = $

In a triangle,if the ex-radii $r_1, r_2, r_3$ are in the ratio $1: 2: 3$,then its sides are in the ratio

If the angles $A, B$ and $C$ of a triangle are in $A.P.$ and if $a, b$ and $c$ denote the length of the sides opposite to $A, B$ and $C$ respectively,then the value of $\frac{a}{b} \sin 2B + \frac{b}{a} \sin 2A$ is