Question

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

• A. $\text{H.P.}$
• B. $\text{A.P.}$
• C. $\text{G.P.}$
• D. None of the above

Answer 1

The correct option is A $\text{H.P.}$

${\sin }^2\frac{A}{2},{\sin }^2\frac{B}{2},{\sin }^2\frac{C}{2}$ are in $\text{H.P.}$
$\Rightarrow \frac{1}{{\sin }^2\frac{A}{2}},\frac{1}{{\sin }^2\frac{B}{2}},\frac{1}{{\sin }^2\frac{C}{2}}$ are in $\text{A.P.}$
$\Rightarrow \frac{1}{{\sin }^2\frac{C}{2}}-\frac{1}{{\sin }^2\frac{B}{2}}=\frac{1}{{\sin }^2\frac{B}{2}}-\frac{1}{{\sin }^2\frac{A}{2}}$
$\Rightarrow \frac{ab}{(s-a)(s-b)}-\frac{ac}{(s-a)(s-c)}=\frac{ac}{(s-a)(s-c)}-\frac{bc}{(s-b)(s-c)}$
$\Rightarrow \frac{a}{s-a}\left(\frac{b(s-c)-c(s-b)}{(s-b)(s-c)}\right)=\frac{c}{s-c}\left(\frac{a(s-b)-b(s-a)}{(s-a)(s-b)}\right)$
$\Rightarrow ab-ac=ac-bc$
$\Rightarrow ab+bc=2ac\Rightarrow \frac{1}{c}+\frac{1}{a}=\frac{2}{b}$
Hence, $a,b,c$ are in $\text{H.P.}$