If in a triangle $ABC$,$a^2+2bc-(b^2+c^2)=ab \sin \frac{C}{2} \cos \frac{C}{2}$,then $\cot (B+C)=$

Consider an obtuse-angled triangle $ABC$ in which the difference between the largest and the smallest angle is $\frac{\pi}{2}$ and whose sides are in arithmetic progression. Suppose that the vertices of this triangle lie on a circle of radius $1$.
$(1)$ Let $a$ be the area of the triangle $ABC$. Then the value of $(64 a)^2$ is
$(2)$ The inradius of the triangle $ABC$ is

With usual notations in $\triangle ABC$,if $\angle B = \frac{\pi}{2}$,and $\tan \frac{A}{2}, \tan \frac{C}{2}$ are roots of the equation $px^2 + qx + r = 0$,$p \neq 0$,then:

If triangle $ABC$ is right-angled at $A$ and $\tan \frac{B}{2}, \tan \frac{C}{2}$ are roots of the equation $ax^2 + bx + c = 0$ $(a \neq 0)$,then:

In $\triangle ABC$,if $\sin^2 A + \sin^2 B = \sin^2 C$ and $l(AB) = 10$,then the maximum value of the area of $\triangle ABC$ is

With usual notations,in $\Delta ABC$,if $b \cos ^{2} \frac{C}{2}+c \cos ^{2} \frac{B}{2}=\frac{3 a}{2}$,then