Q:

A question on trigonometry and solution of triangles

Sides $a, b, c$ of the triangle ABC are in H.P, then prove that $\csc A (\csc A +\cot A); \csc  B (\csc  + \cot B) $& $\csc  C (\csc  C + \cot C)$ are in AP.

I recently came across the above question which I'm not able to solve. I'm not getting any ideas regarding this question. I'm also not able to understand how to algebraically manipulate the expression and use the condition appropriately. Would someone please help me to solve this problem? 
Thanks for help. 

A:

Hint:
$$\csc A(\csc A+\cot A)=\dfrac{1+\cos A}{\sin^2A}=\dfrac1{1-\cos A}$$
$$=\dfrac{2bc}{4(s-b)(s-c)}$$
$$=\dfrac{abc}{2(s-a)(s-b)(s-c)}\cdot\left(\dfrac sa-1\right)$$

