Triangle sides $a,b,c$ are in arithmetic progression. Show $\sin^2(A/2)\csc2A$, $\sin^2(B/2)\csc2B$, $\sin^2(C/2)\csc2C$ are in harmonic progression 
If sides $a,b,c$ of the triangle ABC are in arithmetic progression, prove that $\sin^2\frac{A}{2}\mathrm{cosec}(2A),\sin^2\frac{B}{2}\mathrm{cosec}(2B),\sin^2\frac{C}{2}\mathrm{cosec}(2C)$ are in harmonic progression.

My attempt is as follows:-
$$T_1=\dfrac{1-\cos A}{2\sin2A}$$
$$T_1=\dfrac{1}{4}\dfrac{\sec A-1}{\sin A}$$
$$T_1=\dfrac{2R}{4}\left(\dfrac{\dfrac{2bc}{b^2+c^2-a^2}-1}{a}\right)$$
$$T_1=\dfrac{R}{2}\left(\dfrac{\dfrac{2bc-b^2-c^2+a^2}{b^2+c^2-a^2}}{a}\right)$$
$$T_1=\dfrac{R}{2}\left(\dfrac{\dfrac{a^2-(b-c)^2}{b^2+c^2-a^2}}{a}\right)$$
$$T_1=\dfrac{R}{2}\left(\dfrac{(a+c-b)(a+b-c)}{(b^2+c^2-a^2)a}\right)$$
By symmetry, we can say $T_2=\dfrac{R}{2}\left(\dfrac{(b+c-a)(b+a-c)}{(a^2+c^2-b^2)b}\right)$
$T_3=\dfrac{R}{2}\left(\dfrac{(c+a-b)(c+b-a)}{(a^2+b^2-c^2)c}\right)$
For $T_1,T_2,T_3$ to be in HP, $\dfrac{1}{T_1},\dfrac{1}{T_2},\dfrac{1}{T_3}$ should be in A.P
$$\dfrac{1}{T_1}+\dfrac{1}{T_3}-\dfrac{2}{T_2}=\dfrac{2}{R}\left(\dfrac{(b^2+c^2-a^2)a}{(a+c-b)(a+b-c)}+\dfrac{(a^2+b^2-c^2)c}{(c+a-b)(c+b-a)}-\dfrac{2(a^2+c^2-b^2)b}{(b+c-a)(a+b-c)}\right)$$
$$\dfrac{1}{T_1}+\dfrac{1}{T_3}-\dfrac{2}{T_2}=\dfrac{2}{R}\left(\dfrac{a(b^2+c^2-a^2)(b+c-a)+c(a^2+b^2-c^2)(a+b-c)-2b(a^2+c^2-b^2)(c+a-b)}{(b+c-a)(a+b-c)(c+a-b)}\right)$$
$$\dfrac{1}{T_1}+\dfrac{1}{T_3}-\dfrac{2}{T_2}=\dfrac{2}{R}\left(\dfrac{a(b^3+b^2c-ab^2+bc^2+c^3-ac^2-a^2b-a^2c+a^3)+c(a^3+a^2b-a^2c+b^2a+b^3-b^2c-c^2a-c^2b+c^3)-2b(a^2c+a^3-a^2b+c^3+ac^2-bc^2-b^2c-ab^2+b^3)}{(b+c-a)(a+b-c)(c+a-b)}-\right)$$
$$\dfrac{1}{T_1}+\dfrac{1}{T_3}-\dfrac{2}{T_2}=\dfrac{2}{R}\left(\dfrac{a^4+c^4-2b^4+ab^3-a^3b+ac^3-a^3c+a^3c+b^3c-ac^3-bc^3+ab^2c+abc^2+a^2bc+ab^2c-2a^2bc-2abc^2\cdot\cdot}{(b+c-a)(a+b-c)(c+a-b)}\right)$$
It was getting very difficult to solve from here, is there any other method in which we can solve this question?
 A: $$f(x)=\dfrac{2\sin2x}{1-\cos x}=\dfrac{4\sin x(1-(1-\cos x))}{1-\cos x}=\dfrac{4\sin x}{1-\cos x}-4\sin x=4\cot\dfrac x2-4\sin x$$
Using https://en.m.wikibooks.org/wiki/Trigonometry/Solving_triangles_by_half-angle_formulae and 
https://en.m.wikipedia.org/wiki/Law_of_sines
$f(A)=\dfrac{4s(s-a)}\triangle -\dfrac{4a}{2R}$ where $2s=a+b+c$ and $R$ is the circumradius
Observe that both $s-y$ hence $\dfrac{4s(s-y)}\triangle$ and $\dfrac{4y}{2R}$ will separately form arithmetic progression for $y=a,b,c$ if $a,b,c$ are so.
A: You have not used the fact that $a$, $b$, $c$ are in arithmetic progression. WLOG write $a=b-d$, $c=b+d$; then
\begin{align*}
\dfrac{1}{T_1}+\dfrac{1}{T_3}-\dfrac{2}{T_2}&=\dfrac{2}{R}\left(\dfrac{(b^2+c^2-a^2)a}{(a+c-b)(a+b-c)}+\dfrac{(a^2+b^2-c^2)c}{(c+a-b)(c+b-a)}-\dfrac{2(a^2+c^2-b^2)b}{(b+c-a)(a+b-c)}\right)\\
&= \frac{2}{R}\left(\frac{b^2+3 b d-4 d^2}{b-2 d}+\frac{b^2-3bd-4d^2}{b+2d}-\frac{2b^3+4bd^2}{(b-2d)(b+2d)}\right).
\end{align*}
To see that the parenthesized expression is zero, just compute the combined numerator:
$$(b^2+3bd-4d^2)(b+2d)+(b^2-3bd-4d^2)(b-2d)-2b^3-4b d^2=0.$$
A: The arithmetic progression of sides $b-a = c-b$ leads to
$$\sin B - \sin A = \sin C - \sin B\tag 1$$
With $A+B+C=\pi$,
$$\sin\frac{C}2\sin\frac{B-A}2=\sin\frac{A}2\sin\frac{C-B}2\implies
\frac{\sin\frac{B-A}2}{\sin\frac A2} 
= \frac{\sin\frac{C-B}2}{\sin\frac C2}\tag 2$$
Next, let $I_X = {\sin^2\frac{X}{2}\csc2X}$ to examine
$$\frac1{I_{B}}-\frac1{I_A}=\frac{\sin2B}{\sin^2\frac{B}{2}}-\frac{\sin2A}{\sin^2\frac{A}{2}}
=4\cdot\frac{\cos B\cos\frac{B}{2}\sin\frac{A}{2}-\cos A\cos\frac A2\sin\frac{B}{2}}{\sin\frac{B}{2}\sin\frac{A}{2}}$$
Use the identity $\cos x = 1-2\sin^2\frac x2$ to simplify,
$$\frac14 (\frac1{I_{B}}-\frac1{I_A})= \frac{\sin\frac{A-B}{2}}{\sin\frac{B}{2}\sin\frac{A}{2}}
-(\sin B - \sin A)\tag 3$$
Likewise,
$$\frac14 (\frac1{I_{C}}-\frac1{I_B})= \frac{\sin\frac{B-C}{2}}{\sin\frac{C}{2}\sin\frac{B}{2}}
-(\sin C - \sin B)\tag 4$$
Apply the results (1)-(2) to (3)-(4) to obtain,
$$  \frac1{I_{B}}-\frac1{I_A}=\frac1{I_{C}}-\frac1{I_B}$$
Hence, $I_A $, $I_{B}$ and $I_{C}$ are in harmonic progression.
