Question

If ${\sin }^2\frac{A}{2}{\sin }^2\frac{B}{2}{\sin }^2\frac{C}{2}$ are in $H.P.$, where $a,b,c$ are sides of triangle,then show that $a,b,c$ are in $H.p$.

Answer 1

$\begin{array}{c}Giventhat\\ {\sin }^2\frac{A}{2},{\sin }^2\frac{B}{2},{\sin }^2\frac{C}{2}areinH.P\\ \Rightarrow \frac{1}{{\sin }^2\frac{A}{2}},\frac{1}{{\sin }^2\frac{B}{2}},\frac{1}{{\sin }^2\frac{C}{2}}willbeinA.P\\ \Rightarrow \frac{bc}{\left(s-b\right)\left(s-c\right)},\frac{ac}{s\left(s-a\right)\left(s-b\right)},\frac{ab}{s\left(s-a\right)\left(s-b\right)}willbeinA.P\\ \Rightarrow \frac{bc\left(s-a\right)}{s\left(s-a\right)\left(s-b\right)\left(s-c\right)},\frac{ac\left(s-b\right)}{s\left(s-a\right)\left(s-b\right)\left(s-c\right)},\frac{ab\left(s-c\right)}{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}willbeinA.P\\ \Rightarrow \frac{bc\left(s-a\right)}{{\Delta }^2},\frac{ac\left(s-b\right)}{{\Delta }^2},\frac{ab\left(s-c\right)}{{\Delta }^2}areinA.P\left[\therefore areaof\Delta =\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}\right]\\ \Rightarrow bc\left(s-a\right),ac\left(s-b\right),ab\left(s-c\right)areinA.P\\ \Rightarrow bcs-abc,acs-abc,abs-abcareinA.P\\ \Rightarrow bcs,acsandabsareinA.P\\ Nowdividingbyabcsweget\\ \frac{bcs}{abcs},\frac{acs}{abcs},\frac{abs}{abcs}\\ =\frac{1}{a},\frac{1}{b},\frac{1}{c}areinA.P\\ \therefore a,b,careinH.Pwherea,bandcaresidesofthetriangle.\end{array}$