If $a, \; b, \; c$ are in H.P, prove that $\sin^2 \left(\dfrac{A}{2}\right)$, $\;$ $\sin^2 \left(\dfrac{B}{2}\right)$, $\;$ $\sin^2 \left(\dfrac{C}{2}\right)$ are also in H.P.
Given: $\;$ $a, \; b, \; c$ are in H.P.
$\implies$ $\dfrac{2}{b} = \dfrac{1}{a} + \dfrac{1}{c}$ $\;\;\; \cdots \; (1)$
To prove that (T.P.T) $\;$ $\sin^2 \left(\dfrac{A}{2}\right)$, $\;$ $\sin^2 \left(\dfrac{B}{2}\right)$, $\;$ $\sin^2 \left(\dfrac{C}{2}\right)$ are in H.P.
i.e. $\;$ $T.P.T$ $\;$ $\dfrac{2}{\sin^2 \left(\dfrac{B}{2}\right)} = \dfrac{1}{\sin^2 \left(\dfrac{A}{2}\right)} + \dfrac{1}{\sin^2 \left(\dfrac{C}{2}\right)}$
In $\triangle ABC$,
$\sin \left(\dfrac{A}{2}\right) = \sqrt{\dfrac{\left(s - b\right) \left(s - c\right)}{bc}}$, $\;$ $\sin \left(\dfrac{B}{2}\right) = \sqrt{\dfrac{\left(s - c\right) \left(s - a\right)}{ca}}$, $\;$ $\sin \left(\dfrac{C}{2}\right) = \sqrt{\dfrac{\left(s - a\right) \left(s - b\right)}{ab}}$
where $\;$ $s = \dfrac{a + b + c}{2}$ $\;$ is the semi perimeter of $\triangle ABC$.
$\implies$ $2 s = a + b + c$ $\;\;\; \cdots \; (2)$
$\begin{aligned}
\therefore \; \dfrac{1}{\sin^2 \left(\dfrac{A}{2}\right)} + \dfrac{1}{\sin^2 \left(\dfrac{C}{2}\right)} & = \dfrac{bc}{\left(s - b\right) \left(s - c\right)} + \dfrac{ab}{\left(s - a\right) \left(s - b\right)} \\\\
& = \dfrac{b}{\left(s - b\right)} \left[\dfrac{c}{s - c} + \dfrac{a}{s - a}\right] \\\\
& = \dfrac{b}{\left(s - b\right)} \left[\dfrac{cs - ca + as - ca}{\left(s - c\right) \left(s - a\right)}\right] \\\\
& = \dfrac{b s \left(c + a\right) - 2 abc}{\left(s - a\right) \left(s - b\right) \left(s - c\right)} \;\;\; \cdots \; (3)
\end{aligned}$
We have from equation $(1)$, $\;$ $a + c = \dfrac{2 a c}{b}$
Substituting the value of $\left(a + c\right)$ in equation $(3)$, we have,
$\begin{aligned}
\dfrac{1}{\sin^2 \left(\dfrac{A}{2}\right)} + \dfrac{1}{\sin^2 \left(\dfrac{C}{2}\right)} & = \dfrac{bs \times \dfrac{2 a c}{b} - 2 abc}{\left(s - a\right) \left(s - b\right) \left(s - c\right)} \\\\
& = \dfrac{2s \times ac - 2abc}{\left(s - a\right) \left(s - b\right) \left(s - c\right)} \\\\
& = \dfrac{\left(a + b + c\right)a c - 2 a b c}{\left(s - a\right) \left(s - b\right) \left(s - c\right)} \;\; \left[\text{by equation (2)}\right] \\\\
& = \dfrac{ac \left(a + b + c - 2 b\right)}{\left(s - a\right) \left(s - b\right) \left(s - c\right)} \\\\
& = \dfrac{ac \left(a + c - b\right)}{\left(s - a\right) \left(s - b\right) \left(s - c\right)} \;\;\; \cdots \; (4)
\end{aligned}$
We have from equation $(2)$, $\;$ $a + c = 2 s - b$
Substituting the value of $\left(a + c\right)$ in equation $(4)$, we have,
$\begin{aligned}
\dfrac{1}{\sin^2 \left(\dfrac{A}{2}\right)} + \dfrac{1}{\sin^2 \left(\dfrac{C}{2}\right)} & = \dfrac{a c \left(2 s - b - b\right)}{\left(s - a\right) \left(s - b\right) \left(s - c\right)} \\\\
& = \dfrac{2 a c \left(s - b\right)}{\left(s - a\right) \left(s - b\right) \left(s - c\right)} \\\\
& = 2 \times \dfrac{ac}{\left(s - a\right) \left(s - c\right)}
\end{aligned}$
i.e. $\;$ $\dfrac{1}{\sin^2 \left(\dfrac{A}{2}\right)} + \dfrac{1}{\sin^2 \left(\dfrac{C}{2}\right)} = \dfrac{2}{\sin^2 \left(\dfrac{B}{2}\right)}$
$\implies$ $\sin^2 \left(\dfrac{A}{2}\right)$, $\;$ $\sin^2 \left(\dfrac{B}{2}\right)$, $\;$ $\sin^2 \left(\dfrac{C}{2}\right)$ are in H.P.
Hence proved.