In triangle ABC, if sin2⁡A2,sin2⁡B2 and sin2⁡C2 are in H. P., then a. b and c wiII be in

1sin2⁡A2,1sin2⁡B2,1sin2⁡C2 are in A.P. ⇒1sin2⁡C2−1sin2⁡B2=1sin2⁡B2−1sin2⁡A2⇒ab(s−a)(s−b)−ac(s−a)(s−c)=ac(s−a)(s−c)−bc(s−b)(s−c)⇒as−ab(s−c)−c(s−b)(s−b)(s−c)=cs−ca(s−b)−b(s−a)(s−a)(s−b)⇒ab−ac=ac−bc⇒ab+bc=2ac⇒1c+1a=2b, i.e., a,b,c are in H.P.

In triangle ABC, if sin2⁡A2,sin2⁡B2 and sin2⁡C2 are in H. P., then a. b and c wiII be in

1sin2⁡A2,1sin2⁡B2,1sin2⁡C2 are in A.P. ⇒1sin2⁡C2−1sin2⁡B2=1sin2⁡B2−1sin2⁡A2⇒ab(s−a)(s−b)−ac(s−a)(s−c)=ac(s−a)(s−c)−bc(s−b)(s−c)⇒as−ab(s−c)−c(s−b)(s−b)(s−c)=cs−ca(s−b)−b(s−a)(s−a)(s−b)⇒ab−ac=ac−bc⇒ab+bc=2ac⇒1c+1a=2b, i.e., a,b,c are in H.P.

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

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Detailed Solution

$\begin{array}{l} \frac{1}{\sin^{2} \frac{A}{2}}, \frac{1}{\sin^{2} \frac{B}{2}}, \frac{1}{\sin^{2} \frac{C}{2}} are in A.P. \\ \Rightarrow \frac{1}{\sin^{2} \frac{C}{2}} - \frac{1}{\sin^{2} \frac{B}{2}} = \frac{1}{\sin^{2} \frac{B}{2}} - \frac{1}{\sin^{2} \frac{A}{2}} \\ \Rightarrow \frac{ab}{(s - a) (s - b)} - \frac{ac}{(s - a) (s - c)} = \frac{ac}{(s - a) (s - c)} - \frac{bc}{(s - b) (s - c)} \end{array}$
$\begin{array}{l} \Rightarrow (\frac{a}{s - a}) (\frac{b (s - c) - c (s - b)}{(s - b) (s - c)}) = (\frac{c}{s - c}) (\frac{a (s - b) - b (s - a)}{(s - a) (s - b)}) \\ \Rightarrow ab - ac = ac - bc \\ \Rightarrow ab + bc = 2 ac \\ \Rightarrow \frac{1}{c} + \frac{1}{a} = \frac{2}{b}, i.e., a, b, c are in H.P. \end{array}$