Question

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

• A. $A.P.$
• B. $G.P.$
• C. $H.P.$
• D. $A.G.P.$

Answer 1

The correct option is C $H.P.$

As a,b,c are in H.P $\Rightarrow \frac{1}{b}-\frac{1}{a}=\frac{1}{a}-\frac{1}{b}...\left(1\right)$

By sine rule ,(1)

$\frac{sinA-sinB}{sinAsinB}=\frac{sinB-sinC}{sinBsinC}$

$\Rightarrow \frac{2sin\frac{A-B}{2}cos\frac{A+B}{2}}{2sinA/2cosA/2}=\frac{2sin\frac{B-C}{2}cos\frac{B+C}{2}}{2sinC/2cosC/2}$

$\Rightarrow sin\frac{A-B}{2}cos[\frac{\pi }{2}-\frac{C}{2}]sin\frac{C}{2}cos\frac{C}{2}=sin\frac{B-C}{2}cos(\frac{\pi }{2}-\frac{A}{2})sin\frac{A}{2}cos\frac{A}{2}$

$\Rightarrow sin\frac{A-B}{2}si{n}^2\frac{C}{2}cos\frac{C}{2}=sin\frac{B-C}{2}si{n}^2\frac{A}{2}cos\frac{A}{2}$

$\Rightarrow sin\frac{A-B}{2}sin\frac{A+B}{2}si{n}^2\frac{C}{2}=sin\frac{B-C}{2}sin\frac{B+C}{2}si{n}^2\frac{A}{2}$

$\Rightarrow si{n}^2\frac{C}{2}(si{n}^2\frac{A}{2}-si{n}^2\frac{B}{2})=si{n}^2\frac{A}{2}(si{n}^2\frac{B}{2}-si{n}^2\frac{C}{2})$

Dividing by $si{n}^{2}A/2si{n}^{2}B/2si{n}^{2}C/2$

$\therefore \frac{1}{si{n}^{2}B/2}-\frac{1}{si{n}^{2}A/2}=\frac{1}{si{n}^{2}C/2}-\frac{1}{si{n}^{2}B/2}$

$\Rightarrow si{n}^{2}A/2,si{n}^{2}B/2,si{n}^{2}C/2$ are in H.P

$\therefore$ option C is correct.