Prove that $\sin^2\frac{A}{2}\csc2A$, $\sin^2\frac{B}{2}\csc2B$, $\sin^2\frac{C}{2}\csc2C$ are in harmonic progression 
If sides $a,b,c$ of $\triangle ABC$ are in arithmetic progression (AP), then prove that 
  $$\sin^2\frac{A}{2}\csc2A, \quad\sin^2\frac{B}{2}\csc2B, \quad \sin^2\frac{C}{2}\csc2C$$ are in harmonic progression (HP).


In order for 
$$\sin^2\frac{A}{2}\csc2A, \quad \sin^2\frac{B}{2}\csc2B, \quad \sin^2\frac{C}{2}\csc2C$$ 
to be in HP, their reciprocals 
$$\frac{\sin 2A}{\sin^2\frac{A}{2}}, \quad \frac{\sin 2B}{\sin^2\frac{B}{2}}, \quad \frac{\sin 2C}{\sin^2\frac{C}{2}}$$
need to be in AP. That is,
$$2\frac{\sin 2B}{\sin^2\frac{B}{2}}=\frac{\sin 2A}{\sin^2\frac{A}{2}}+\frac{\sin 2C}{\sin^2\frac{C}{2}} \tag{1}$$
We are given that $2b=a+c$; thus, by the Sine Rule,
$$2\sin B=\sin A+\sin C \tag{2}$$
I do not know how to prove equation $(1)$ from equation $(2)$.

$$\dfrac1{\sin^2\frac{A}{2}\csc2A}=\dfrac{2\sin A\left(1-2\sin^2\frac{A}{2}\right)}{\sin^2\frac{A}{2}}=4\cot\dfrac A2-4\sin A$$
So, we need to prove  $\cot\dfrac A2$ etc. are in A.P.
Now use $\cot\dfrac A2=\dfrac{s(s-a)}{\triangle}$
