In Delta ABC, if sin^2 frac{A}{2}, sin^2 frac | vedclass

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(C) Given that $\sin^2 \frac{A}{2}, \sin^2 \frac{B}{2}, \sin^2 \frac{C}{2}$ are in $H.P.$Therefore,$\frac{1}{\sin^2 \frac{A}{2}}, \frac{1}{\sin^2 \fra

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(C) Given that $\sin^2 \frac{A}{2}, \sin^2 \frac{B}{2}, \sin^2 \frac{C}{2}$ are in $H.P.$Therefore,$\frac{1}{\sin^2 \frac{A}{2}}, \frac{1}{\sin^2 \frac{B}{2}}, \frac{1}{\sin^2 \frac{C}{2}}$ are in $A.P.$Using the formula $\sin^2 \frac{A}{2} = \frac{(s-b)(s-c)}{bc},$ we have $\frac{bc}{(s-b)(s-c)}, \frac{ac}{(s-a)(s-c)}, \frac{ab}{(s-a)(s-b)}$ are in $A.P.$This implies $\frac{ac}{(s-a)(s-c)} - \frac{bc}{(s-b)(s-c)} = \frac{ab}{(s-a)(s-b)} - \frac{ac}{(s-a)(s-c)}.$Simplifying this expression leads to $ab + bc = 2ac.$Dividing by $abc,$ we get $\frac{1}{c} + \frac{1}{a} = \frac{2}{b}.$This shows that $a, b, c$ are in $H.P.$

In $\Delta ABC,$ if $\sin^2 \frac{A}{2}, \sin^2 \frac{B}{2}, \sin^2 \frac{C}{2}$ are in $H.P.,$ then $a, b, c$ will be in

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